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CBSE NCERT NOTES CLASS 12 PHYSICS CHAPTER 2 ELECTROSTATICS POTENTIAL AND CAPACITANCE

Electric Potential Energy

Potential Energy of charge q at a point (in the presence of field due to any charge configuration) is the work done by the external force (equal and opposite to the electric force) in bringing the charge q from infinity to that point without accelerating,

We can define electric potential energy difference between two points as the work required to be done by an external force in moving (without accelerating) charge q from one point to another in electric field of any arbitrary charge configuration.

Consider an electrostatic field E due to a charge Q placed at the origin. Now, bring a test charge q from a point R to a point P against the repulsive force on it due to the charge Q. This will happen if Q and q are either positive or both negative.

Let us take both Q, q > 0. Following conditions are assumed,

1. The test charge q is so small that it does not disturb the original configuration.


2. Second, in bringing the charge q from R to P, the applied external force F_ext is just enough to counter the repulsive electric force F_E (i.e, F_ext= –F_E). This means there is no net force on or acceleration of the charge q when it is brought from R to P, i.e., it is brought with infinitesimally slow constant speed.

Work done by the external force is the negative of the work done by the electric force, and gets fully stored in the form of potential energy of the charge q.

If the external force is removed on reaching P, the electric force will take the charge away from Q, and the stored energy (potential energy) at P is used to provide kinetic energy to the charge q in such a way that the sum of the kinetic and potential energies is constant.

Thus, work done by external force in moving a charge q from R to P is

$${\mathrm{W}}_{\mathrm{R}\mathrm{P}}={\int }_{\mathrm{R}}^{\mathrm{P}}{\mathrm{F}}_{\mathrm{e}\mathrm{x}\mathrm{t}}.\mathrm{d}\mathrm{r}={\int }_{\mathrm{R}}^{\mathrm{P}}{\mathrm{F}}_{\mathrm{E}}.\mathrm{d}\mathrm{r}$$

At every point in electric field, a particle with charge q possesses a certain electrostatic potential energy, this work done increases its potential energy by an amount equal to potential energy difference between points R and P.

Thus, potential energy difference will be,

ΔU = U_P − U_R = W_RP.

Since this displacement is in an opposite direction to the electric force, the work done by electric field is negative, i.e., –W_RP.

1. Work done by an electrostatic field in moving a charge from one point to another depends only on the initial and the final points and is independent of the path taken to go from one point to the other. This is the fundamental characteristic of a conservative force.

   Mathematically,

   $$\oint \mathrm{E}.\mathrm{d}\mathrm{l}=0$$



2. We normally take electrostatic potential energy zero at infinity, i.e., if R is at infinity,

   W_∞P = U_P – U_∞

Electrostatic Potential

Electrostatic Potential at any point is the work done in bringing a unit +ve charge from infinity to that point.

Potential Difference between two points (say R and P) is the work done in bringing a unit positive charge from point R to P

$$=\mathrm{ }{\mathrm{V}}_{\mathrm{P}}\mathrm{ }–\mathrm{ }{\mathrm{V}}_{\mathrm{R}}\mathrm{ }=\frac{{\mathrm{U}}_{\mathrm{P}}\mathrm{ }–\mathrm{ }{\mathrm{U}}_{\mathrm{R}}}{\mathrm{q}},$$

• SI Unit is Volt or V = J/C.

Potential due to a point Charge

Consider a point charge Q at the origin. For Q > 0, the work done against the repulsive force on the test charge is positive. Choose a convenient path along the radial direction from infinity to the point P.

At some intermediate point P′ on the path, the electrostatic force on a unit positive charge is

$$\frac{\mathrm{Q}}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}{{\mathrm{r}}^{\mathrm{\text{'}}}}^2}\widehat{\mathrm{r}}\mathrm{\text{'}}$$

where ${\displaystyle \widehat{\mathrm{r}}\mathrm{\text{'}}}$ is the unit vector along OP′.

Work done against this force from r′ to r′ + ∆r′ is

$$\mathrm{\Delta }\mathrm{W}=-\mathrm{ }\frac{\mathrm{Q}}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}{{\mathrm{r}}^{\mathrm{\text{'}}}}^2}\mathrm{\Delta }\mathrm{r}\mathrm{\text{'}}$$

The negative sign appears because for ∆r′ < 0, therefore ∆W is positive.

Total work done (W) by the external force can be obtained by integrating from r′ = ∞ to r′ = r,

$$\mathrm{W}=-{\int }_{\mathrm{\infty }}^{\mathrm{r}}\mathrm{ }\frac{\mathrm{Q}}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}{{\mathrm{r}}^{\mathrm{\text{'}}}}^2}\mathrm{d}{\mathrm{r}}^{\mathrm{\text{'}}}$$

$$={\left|\frac{\mathrm{Q}}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}{\mathrm{r}}^{\mathrm{\text{'}}}}\right|}_{\mathrm{\infty }}^{\mathrm{r}}$$

$$=\mathrm{ }\frac{\mathrm{Q}}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{r}}\mathrm{ }$$

This, by definition is the potential at P due to the charge Q or

$$\mathrm{V(r)}=\mathrm{ }\frac{\mathrm{Q}}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{r}}$$

We see that, potential is inversely proportional to r and electric field is inversely proportional to square of r.

$$\mathrm{V}\mathrm{ }\propto \frac{1}{\mathrm{r}}$$

And

$$\mathrm{E}\mathrm{ }\propto \frac{1}{{\mathrm{r}}^2}.\mathrm{ }$$

The graph of V and E with respect to distance r can be plotted as shown in the figure.

Potential due to a System of Charges

Consider a system of charges q₁, q₂,…, q_n at disances ${\displaystyle {\mathrm{r}}_{1\mathrm{P}}}$, ${\displaystyle {\mathrm{r}}_{2\mathrm{P}}}$,…, ${\displaystyle {\mathrm{r}}_{\mathrm{n}\mathrm{P}}}$ from a point P. The potential V₁ at P due to the charge q₁ is

$${\mathrm{V}}_1=\mathrm{ }\frac{{\mathrm{q}}_1}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}{\mathrm{r}}_{1\mathrm{P}}}$$

 where r_1P is the distance between q₁  and P.

Similarly, the potential V₂ due to q₂ and V₃ due to q₃ are given by

$${\mathrm{V}}_2=\mathrm{ }\frac{{\mathrm{q}}_2}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}{\mathrm{r}}_{2\mathrm{P}}}$$

$${\mathrm{V}}_3=\mathrm{ }\frac{{\mathrm{q}}_3}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}{\mathrm{r}}_{3\mathrm{P}}}$$

where r_2P and r_3P are the distances of P from charges q₂ and q₃, respectively; and so on for the potential due to other charges.

Since potential is related to the work done by the field, electrostatic potential also follows the superposition principle. By the superposition principle, the potential V at P due to the total charge configuration is the algebraic sum of the potentials due to the individual charges

V = V₁ + V₂ + ... + V_n

$$\mathrm{V}=\mathrm{ }\frac{1}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}}\left(\frac{{\mathrm{q}}_1}{{\mathrm{r}}_{1\mathrm{P}}}+\frac{{\mathrm{q}}_2}{{\mathrm{r}}_{2\mathrm{P}}}+\mathrm{ }\dots .\mathrm{ }+\frac{{\mathrm{q}}_{\mathrm{n}}}{{\mathrm{r}}_{\mathrm{n}\mathrm{P}}}\right)$$

Electric potential due to a Spherical shell

• The electric field outside a spherical shell of radius R is, as if the entire charge is concentrated at the centre. Thus, the potential outside the shell (r ≥ R) is given by

  $$\mathrm{V}=\mathrm{ }\frac{\mathrm{q}}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{r}}$$

  where q is the total charge on the shell and R its radius.

  Proof

  Electric charge on the shell,

  Q = σ A = 4π σ R²

  Electric field at r > R,

  $$\mathrm{E}\mathrm{ }=\frac{1}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}}\frac{\mathrm{Q}}{{\mathrm{r}}^2}$$

  Electric potential at r > R

  $$\mathrm{V}\mathrm{ }=\mathrm{ }-{\int }_{\mathrm{\infty }}^{\mathrm{r}}\frac{1}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}}\frac{\mathrm{Q}}{{\mathrm{r}}^2}\mathrm{ }\mathrm{d}\mathrm{r}$$

  $$=\mathrm{ }\frac{1}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}}\frac{\mathrm{Q}}{\mathrm{r}}$$



• The electric field inside the shell is zero. This implies that potential is constant inside the shell (as no work is done in moving a charge inside the shell), and, therefore, equals its value at the surface, which is

  $$\mathrm{V}=\mathrm{ }\frac{\mathrm{q}}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{R}}$$

  Proof:

  Electric field at r < R, E = 0

  Electric potential at r < R

  $$\mathrm{V}\mathrm{ }=\mathrm{ }-{\int }_{\mathrm{\infty }}^{\mathrm{r}}\mathrm{E}\mathrm{ }\mathrm{d}\mathrm{r}$$

  $$=-{\int }_{\mathrm{\infty }}^{\mathrm{R}}\frac{1}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}}\frac{\mathrm{Q}}{{\mathrm{r}}^2}\mathrm{ }\mathrm{d}\mathrm{r}\mathrm{ }$$

  $$=\mathrm{ }\frac{1}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}}\frac{\mathrm{Q}}{\mathrm{R}}$$

The plot of r vs V will be

Electric potential of a uniformly charged sphere of radius a

The electric field at B outside the sphere is, as if the entire charge is concentrated at the centre of the sphere,

$$\overrightarrow{\mathrm{E}}=\frac{\mathrm{Q}}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}{\mathrm{r}}^2}\widehat{\mathrm{r}}$$

Therefore,

$${\mathrm{V}}_{\mathrm{B}}=\mathrm{V}\left(\mathrm{r}\right)-\mathrm{ }\mathrm{V}\left(\mathrm{\infty }\right)$$

$$=\mathrm{ }-{\int }_{\mathrm{\infty }}^{\mathrm{r}}\mathrm{E}\mathrm{ }\mathrm{d}\mathrm{r}$$

$$=\mathrm{ }-{\int }_{\mathrm{\infty }}^{\mathrm{r}}\frac{\mathrm{Q}}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}{\mathrm{r}}^2}\mathrm{ }\mathrm{d}\mathrm{r}$$

$$=\mathrm{ }\frac{1}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}}\frac{\mathrm{Q}}{\mathrm{r}}$$

The electric potential at D inside the sphere is given by

$${\mathrm{V}}_{\mathrm{D}}=\mathrm{V}\left(\mathrm{r}\right)-\mathrm{ }\mathrm{V}\left(\mathrm{\infty }\right)$$

$$=\mathrm{V}\left(\mathrm{r}\right)-\mathrm{V}\left(\mathrm{a}\right)+\mathrm{V}\left(\mathrm{a}\right)-\mathrm{ }\mathrm{V}\left(\mathrm{\infty }\right)$$

$$=\mathrm{ }-{\int }_{\mathrm{\infty }}^{\mathrm{a}}\mathrm{E}\mathrm{ }\mathrm{d}\mathrm{r}-{\int }_{\mathrm{a}}^{\mathrm{r}}\mathrm{E}\mathrm{ }\mathrm{d}\mathrm{r}$$

$$=\mathrm{ }\frac{1}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}}\frac{\mathrm{Q}}{\mathrm{a}}-\frac{1}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}}\frac{\mathrm{Q}}{{\mathrm{a}}^3}\frac{\left({\mathrm{r}}^2-{\mathrm{a}}^2\right)}{2}$$

$$=\mathrm{ }\frac{1}{8\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}}\frac{\mathrm{Q}}{\mathrm{a}}\left(3-\frac{{\mathrm{r}}^2}{{\mathrm{a}}^2}\right)\mathrm{ }$$

We can plot the graph of V wrt r as follows,

Potential due to an Electric Dipole

Take the origin at the centre of the dipole. Since potential is related to the work done by the field, electrostatic potential also follows the superposition principle. Thus, the potential due to the dipole is the sum of potentials due to the charges q and –q

Now electric potential at P due to charge +q

$${\mathrm{V}}_{+\mathrm{q}}\mathrm{ }=\frac{\mathrm{k}\mathrm{q}}{{\mathrm{r}}_1}$$

And

$$\mathrm{ }{\mathrm{V}}_{-\mathrm{q}}\mathrm{ }=-\frac{\mathrm{k}\mathrm{q}}{{\mathrm{r}}_2}$$

And electric potential at P due to charge +q

$$\mathrm{V}={\mathrm{V}}_{+\mathrm{q}}+{\mathrm{V}}_{-\mathrm{q}}=\mathrm{k}\mathrm{q}\left(\frac{1}{{\mathrm{r}}_1}-\frac{1}{{\mathrm{r}}_2}\right)\mathrm{ }$$

where r₁ and r₂ are the distances of the point P from q and –q, respectively.

Now, by geometry, we can see,

$${\mathrm{r}}_1^2={\mathrm{r}}^2+{\mathrm{a}}^2-2\mathrm{a}\mathrm{r}\mathrm{ }\mathrm{cos\; \theta }$$

${\displaystyle \mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{\mathrm{r}}_2^2={\mathrm{r}}^2+{\mathrm{a}}^2+2\mathrm{a}\mathrm{r}\mathrm{ }\mathrm{cos\; \theta }}$

That is,

$${\mathrm{r}}_1^2={\mathrm{r}}^2\left(1+\frac{{\mathrm{a}}^2}{{\mathrm{r}}^2}-\frac{2\mathrm{a}}{\mathrm{r}}\mathrm{cos\; \theta }\right)\mathrm{ }$$

$$\mathrm{\cong }{\mathrm{r}}^2\left(1-\frac{2\mathrm{a}}{\mathrm{r}}\mathrm{cos\; \theta }\right)$$

And

$${\mathrm{r}}_2^2={\mathrm{r}}^2\left(1+\frac{{\mathrm{a}}^2}{{\mathrm{r}}^2}+\frac{2\mathrm{a}}{\mathrm{r}}\mathrm{cos\; \theta }\right)$$

$$\mathrm{\cong }{\mathrm{r}}^2\left(1+\frac{2\mathrm{a}}{\mathrm{r}}\mathrm{cos\; \theta }\right)$$

If r is much greater than a ( r >> a ), we may ignore terms of higher order in ${\displaystyle \frac{\mathrm{a}}{\mathrm{r}}}$

Using the Binomial theorem and retaining terms up to the first order in ${\displaystyle \frac{\mathrm{a}}{\mathrm{r}}}$; we get,

$$\frac{1}{{\mathrm{r}}_1}\mathrm{\cong }\frac{1}{\mathrm{r}}{\left(1-\frac{2\mathrm{a}}{\mathrm{r}\mathrm{ }\mathit{cos\; \theta }}\right)}^{-\frac{1}{2}}$$

$$=\frac{1}{\mathrm{r}}\left(1+\frac{\mathrm{a}}{\mathrm{r}\mathrm{ }\mathit{cos\; \theta }}\right)$$

And

$$\mathrm{ }\frac{1}{{\mathrm{r}}_2}\mathrm{\cong }\frac{1}{\mathrm{r}}{\left(1+\frac{2\mathrm{a}}{\mathrm{r}\mathrm{ }\mathit{cos\; \theta }}\right)}^{-\frac{1}{2}}$$

$$=\mathrm{ }\frac{1}{\mathrm{r}}\left(1-\frac{\mathrm{a}}{\mathrm{r}\mathrm{ }\mathit{cos\; \theta }}\right)$$

Thus

$$\mathrm{V}=\mathrm{k}\mathrm{q}\left(\frac{1}{{\mathrm{r}}_1}-\frac{1}{{\mathrm{r}}_2}\right)$$

$$=\mathrm{ }\mathrm{k}\mathrm{q}\left[\frac{1}{\mathrm{r}}\left(1+\frac{\mathrm{a}}{\mathrm{r}\mathrm{ }\mathit{cos\; \theta }}\right)\right]-\frac{1}{\mathrm{r}}\left(1-\frac{\mathrm{a}}{\mathrm{r}\mathrm{ }\mathrm{ }\mathit{cos\; \theta }}\right)$$

$$=\mathrm{ }\frac{\mathrm{k}\mathrm{ }\mathrm{q}\mathrm{ }2\mathrm{a}\mathrm{ cos\; \theta }}{{\mathrm{r}}^2}=\frac{\mathrm{k}\mathrm{ }\mathrm{p}\mathrm{ cos\; \theta }}{{\mathrm{r}}^2}=\frac{\mathrm{k}\mathrm{ }\vec{\mathrm{p}}.\widehat{\mathrm{r}}}{{\mathrm{r}}^2}$$

where ${\displaystyle \widehat{\mathrm{r}}}$ is the unit vector along ${\displaystyle \overrightarrow{\mathrm{O}\mathrm{P}}}$.

Equation is approximately true only for distances large compared to the size of the dipole, so that higher order terms in ${\displaystyle \frac{\mathrm{a}}{\mathrm{r}}}$ are negligible.

For a point dipole p at the origin,

Potential on the dipole axis (i.e. for, θ = 0, π) is given by,

$$\mathrm{V}=\frac{\mathrm{k}\mathrm{ }\mathrm{p}}{{\mathrm{r}}^2}$$

(Positive sign for θ = 0, negative sign for θ = π.)

The potential due to dipole in the equatorial plane (θ = ${\displaystyle \frac{\mathrm{\pi }}{2}}$) is zero.

1. The potential due to a dipole depends not just on r but also on the angle between the position vector ${\displaystyle \overrightarrow{\mathrm{r}}}$ and the dipole moment vector, ${\displaystyle \overrightarrow{\mathrm{p}}}$.


2. The electric dipole potential falls off, at large distance, as ${\displaystyle \frac{1}{{\mathrm{r}}^2}}$, not as ${\displaystyle \frac{1}{\mathrm{r}}}$, which is characteristic of the potential due to a single charge.

Equipotential Surfaces

An equipotential surface is a surface with a constant value of potential at all points on the surface.

For a single charge q, the potential is given by

$$\mathrm{V(r)}\mathrm{ }=\mathrm{ }\frac{\mathrm{q}}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{r}}$$

This shows that V is a constant if r is constant. Thus, equipotential surfaces of a single point charge are concentric spherical surfaces centred at the charge and electric field lines are radial, starting from the charge if q > 0.

• In general: for any charge configuration, equipotential surface through a point is normal to the electric field at that point.


• No work is done in moving a point charge from one point to another in equipotential surface.


• For a uniform electric field ${\displaystyle \overrightarrow{\mathrm{E}}}$, say, along the x-axis, the equipotential surfaces are planes normal to the x-axis, i.e., planes parallel to the y-z plane. Equipotential surfaces for (a) a dipole and (b) two identical positive charges are shown in figure.

Relation between Electric Field and Potential

Consider two closely spaced equipotential surfaces A and B with potential values V and V + δV, where δV is the change in V in the direction of the electric field E.

Let P be a point on the surface B. δl is the perpendicular distance of the surface A from P.

Imagine that a unit positive charge is moved along this perpendicular from the surface B to surface A against the electric field. The work done in this process is E δl.

This work equals the potential difference V_A –V_B. Thus,

E δl = V − (V + δV)

= – δV

$$\Rightarrow \mathrm{ }\mathrm{ }\mathrm{ }\mathrm{E}=\mathrm{ }-\frac{\mathrm{\delta }\mathrm{V}}{\mathrm{\delta }\mathrm{l}}$$

Since δV is negative,

δV = – | δV |

So, we can write,

$$\left|\mathrm{ }\mathrm{E}\mathrm{ }\right|=\mathrm{ }-\frac{\mathrm{\delta }\mathrm{V}}{\mathrm{\delta }\mathrm{l}}=\mathrm{ }+\frac{\left|\mathrm{\delta }\mathrm{V}\right|}{\mathrm{\delta }\mathrm{l}}$$

Important conclusions

1. Electric field is in the direction in which the potential decreases steepest.


2. Its magnitude is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.

Potential energy of a system of charges

Consider a group of charges q₁, q₂, q₃, …, q_n. Suppose, first the charge q₁ is brought from infinity to the point r₁. There is no external field against which work needs to be done, so work done in bringing q₁ from infinity to r₁ is zero. This charge produces a potential in space given by

$${\mathrm{V}}_1=\mathrm{ }\frac{{\mathrm{q}}_1}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}{\mathrm{r}}_{1\mathrm{P}}}$$

Work done in bringing charge q₂ from infinity to the point r₂ is q₂ times the potential at r₂ due to q₁, that is,

Work done on q₂

$$\mathrm{W}=\mathrm{ }{\mathrm{V}}_1\times {\mathrm{q}}_2=\mathrm{ }\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}{\mathrm{r}}_{12}}$$

where r₁₂ is the distance between points 1 and 2.

Since electrostatic force is conservative, this work gets stored in the form of potential energy of the system. Thus, the potential energy of a system of two charges q₁ and q₂ is

$$\mathrm{U}=\mathrm{ }\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}{\mathrm{r}}_{12}}$$

The charges q₁ and q₂ produce a potential, at point P, which is given by

$${\mathrm{V}}_{\mathrm{1,2}}=\mathrm{ }\frac{1}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}}\left(\frac{{\mathrm{q}}_1}{{\mathrm{r}}_{1\mathrm{P}}}+\frac{{\mathrm{q}}_2}{{\mathrm{r}}_{2\mathrm{P}}}\right)$$

Work done in bringing q₃ from infinity to the point with position vector, ${\displaystyle \overrightarrow{{\mathrm{r}}_3}}$ is,

Work done on q₃

$${{=\mathrm{q}}_3\times \mathrm{V}}_{\mathrm{1,2}}=\mathrm{ }\frac{1}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}}\left(\frac{{\mathrm{q}}_1{\mathrm{q}}_3}{{\mathrm{r}}_{13}}+\frac{{\mathrm{q}}_2{\mathrm{q}}_3}{{\mathrm{r}}_{23}}\right)$$

The total work done in assembling the charges at the given locations is,

$$\mathrm{U}\mathrm{ }=\frac{1}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}}\left(\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}_{12}}+\frac{{\mathrm{q}}_1{\mathrm{q}}_3}{{\mathrm{r}}_{13}}+\frac{{\mathrm{q}}_2{\mathrm{q}}_3}{{\mathrm{r}}_{23}}\right)$$

• The potential energy is characteristic of the present state of configuration, and not the way the state is achieved.

Potential energy of a system of charges in an external field – Two Charges

Consider two charges q₁ and q₂ with position vectors, ${\displaystyle \overrightarrow{{\mathrm{r}}_1}}$ and ${\displaystyle \overrightarrow{{\mathrm{r}}_2}}$ respectively, in an external field.

First, we calculate the work done in bringing the charge q₁ from infinity to a point with position vector ${\displaystyle \overrightarrow{{\mathrm{r}}_1}}$.

Work done in this step = q₁ V(${\displaystyle \overrightarrow{{\mathrm{r}}_1}}$).

Next, we consider the work done in bringing q₂ to to a point with position vector ${\displaystyle \overrightarrow{{\mathrm{r}}_2}}$.

In this step, work is done not only against the external field E but also against the field due to q₁.

Work done on q₂ against the external field = q₂ V(${\displaystyle \overrightarrow{{\mathrm{r}}_2}}$)

and work done on q₂ against the field due to q₁

$$=\mathrm{ }\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}{\mathrm{r}}_{12}}$$

where r₁₂ is the distance between q₁ and q₂.

Therefore, work done in bringing q₂ to r₂

$$=\mathrm{ }{\mathrm{q}}_2\mathrm{V}\left(\overrightarrow{{\mathrm{r}}_2}\right)+\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}{\mathrm{r}}_{12}}$$

Thus, Potential energy of the system = the total work done in assembling the configuration

$$\mathrm{U}\mathrm{ }=\mathrm{ }{\mathrm{q}}_1\mathrm{ }\mathrm{V}\left(\overrightarrow{{\mathrm{r}}_1}\right)+{\mathrm{q}}_2\mathrm{V}\left(\overrightarrow{{\mathrm{r}}_2}\right)+\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}{\mathrm{r}}_{12}}$$

Potential energy of a dipole in an external field

Consider a dipole with charges q₁ = +q and q₂ = –q placed in a uniform electric field E.

In a uniform electric field, the dipole experiences no net force; but experiences a torque ${\displaystyle \overrightarrow{\mathrm{\tau }}}$ given by,

${\displaystyle \overrightarrow{\mathrm{\tau }}\mathrm{ }=\mathrm{ }\overrightarrow{\mathrm{p}}\times \overrightarrow{\mathrm{E}}}$,

which will tend to rotate it (unless ${\displaystyle \overrightarrow{\mathrm{p}}}$ is parallel or antiparallel to ${\displaystyle \overrightarrow{\mathrm{E}}}$).

Suppose an external torque ${\displaystyle {\overrightarrow{\mathrm{\tau }}}_{\mathrm{e}\mathrm{x}\mathrm{t}}}$ is applied in such a manner that it just neutralises this torque and rotates it in the plane of paper from angle θ₀ to angle θ₁ at an infinitesimal small angular speed and without angular acceleration.

The amount of work done by the external torque will be given by

$$\mathrm{W}={\int }_{{\mathrm{\theta }}_{\mathrm{o}}}^{{\mathrm{\theta }}_1}{\mathrm{\tau }}_{\mathrm{e}\mathrm{x}\mathrm{t}}\left(\mathrm{\theta }\right)\mathrm{ }\mathrm{d}\mathrm{\theta }$$

$$=\mathrm{ }{\int }_{{\mathrm{\theta }}_{\mathrm{o}}}^{{\mathrm{\theta }}_1}\mathrm{p}\mathrm{E}\mathrm{ }\mathrm{ }\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{d}\mathrm{\theta }$$

$$={\left[-\mathrm{p}\mathrm{E}\mathit{cos}\right]}_{{\mathrm{\theta }}_{\mathrm{o}}}^{{\mathrm{\theta }}_1}$$

$$=\mathrm{p}\mathrm{E}\mathrm{ }(\mathit{cos}{\mathrm{\theta }}_{\mathrm{o}}-\mathrm{ }\mathit{cos}{\mathrm{\theta }}_1)$$

We can then associate potential energy U(θ) with an inclination θ of the dipole.

Similar to other potential energies, there is a freedom in choosing the angle where the potential energy U is taken to be zero. A natural choice is to take θ₀ = ${\displaystyle \frac{\mathrm{\pi }}{2}}$.

$$\mathrm{U}\left(\mathrm{\theta }\right)=\mathrm{p}\mathrm{E}\mathrm{ }\left(\mathit{cos}\frac{\mathrm{\pi }}{2}-\mathrm{ }\mathit{cos}\mathrm{\theta }\right)$$

$$=\mathrm{ }-\mathrm{p}\mathrm{E}\mathrm{ }\mathrm{ }\mathit{cos}\mathrm{\theta }=-\mathrm{ }\overrightarrow{\mathrm{p}}.\overrightarrow{\mathrm{E}}$$

Electrostatics of conductors

Conductors contain mobile charge carriers. In metallic conductors, these charge carriers are electrons. In a metal, the outer (valence) electrons part away from their atoms and are free to move. These electrons are free within the metal but not free to leave the metal.

In an external electric field, they drift against the direction of the field. The positive ions made up of the nuclei and the bound electrons remain held in their fixed positions. In electrolytic conductors, the charge carriers are both positive and negative ions.

1. Inside a conductor, electrostatic field is zero

   In the static situation, the free charges are so distributed on the surface so that the electric field is zero everywhere inside.

   • Electrostatic field is zero inside a conductor.


2. At the surface of a charged conductor, electrostatic field must be normal to the surface at every point.

   If ${\displaystyle \overrightarrow{\mathrm{E}}}$ were not normal to the surface, it would have some non-zero component along the surface. Free charges on the surface of the conductor would then experience force and move. In the static situation, therefore, ${\displaystyle \overrightarrow{\mathrm{E}}}$ should have no tangential component (that is it should be normal to the surface).

   3. The interior of a conductor can have no excess charge in the static situation.

      A neutral conductor has equal amounts of positive and negative charges in every small volume or surface element. When the conductor is charged, the excess charge can reside only on the surface in the static situation. This follows from the Gauss’s law.

      Consider any arbitrary volume element v inside a conductor. On the closed surface S bounding the volume element v, electrostatic field is zero. Thus the total electric flux through S is zero. Hence, by Gauss’s law, there is no net charge enclosed by S.

   
   
   4. Electrostatic potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface.

      This follows from results 1 and 2 above. Since E = 0 inside the conductor and has no tangential component on the surface, no work is done in moving a small test charge within the conductor and on its surface. That is, there is no potential difference between any two points inside or on the surface of the conductor.



3. Electric field at the surface of a charged conductor

To derive the result, choose a pill box (a short cylinder) as the Gaussian surface about any point P on the surface. The pill box is partly inside and partly outside the surface of the conductor. It has a small area of cross section δS and negligible height.

Just inside the surface, the electrostatic field is zero; just outside, the field is normal to the surface with magnitude E.

Thus, the contribution to the total flux through the pill box comes only from the outside (circular) cross-section of the pill box.

Since over the small area δS, E may be considered constant and E and δS are parallel or antiparallel, the flux equals ± E δS (positive for σ > 0, negative for σ < 0),

The charge enclosed by the pill box is σ δS.

By Gauss’s law

$$\mathrm{E}\mathrm{ }\mathrm{\delta }\mathrm{S}=\frac{\left|\mathrm{\sigma }\right|\mathrm{ }\mathrm{\delta }\mathrm{S}}{{\mathrm{\epsilon }}_{\mathrm{o}}}\mathrm{ }$$

$$\Rightarrow \mathrm{ }\mathrm{E}=\frac{\left|\mathrm{\sigma }\right|}{{\mathrm{\epsilon }}_{\mathrm{o}}}\mathrm{ }\mathrm{ }$$

$$\Rightarrow \mathrm{ }\overrightarrow{\mathrm{E}}=\frac{\left|\mathrm{\sigma }\right|}{{\mathrm{\epsilon }}_{\mathrm{o}}}\widehat{\mathrm{n}}\mathrm{ }$$

6. Electrostatic shielding

   Consider a conductor with a cavity, with no charges inside the cavity, the electric field inside the cavity is zero, whatever be the size and shape of the cavity and whatever be the charge on the conductor and the external fields in which it might be placed.

   • The effect can be made use of in protecting sensitive instruments from outside electrical influence.

Dielectrics and Polarisation

Dielectrics are non-conducting substances. They have no (or negligible number of) charge carriers. When a conductor is placed in an external electric field, the free charge carriers move and charge distribution in the conductor adjusts itself in such a way that the electric field due to induced charges opposes the external field within the conductor.

In a dielectric, this free movement of charges is not possible. The external field induces dipole moment by stretching or re-orienting molecules of the dielectric. The collective effect of all the molecular dipole moments is net charges on the surface of the dielectric which produce a field that opposes the external field. Unlike in a conductor, however, the opposing field so induced does not exactly cancel the external field. It only reduces it. The extent of the effect depends on the nature of the dielectric.

Polarisation in non-polar dielectric

In an external electric field, the positive and negative charges of a nonpolar molecule are displaced in opposite directions. The displacement stops when the external force on the onstituent charges of the molecule is alanced by the restoring force (due to internal fields in the molecule).

Polarisation in polar dielectric

A dielectric with polar molecules also develops a net dipole moment in an external field, but for a different reason.

In the absence of any external field, the different permanent dipoles are oriented randomly due to thermal agitation; so the total dipole moment is zero.

When an external field is applied, the individual dipole moments tend to align with the field. When summed over all the molecules, there is then a net dipole moment in the direction of the external field, i.e., the dielectric is polarised.

The factors affecting polarisation

The extent of polarisation depends on the relative strength of two mutually opposite factors:

• The dipole potential energy in the external field tending to align the dipoles with the field and


• Thermal energy tending to disrupt the alignment.


• There may be, in addition, the ‘induced dipole moment’ effect as for non-polar molecules, but generally the alignment effect is more important for polar molecules.

The dielectric is thus said to be polarised in either case by the external field.

Definition of Polarisation

The dipole moment per unit volume is called polarisation and is denoted by ${\displaystyle \mathrm{P}}$. For linear isotropic dielectrics, P = 𝜒_eE where 𝜒_e is a constant characteristic of the dielectric and is known as the electric susceptibility of the dielectric medium.

How does the polarised dielectric modify the original external field inside it?

Let us consider, for simplicity, a rectangular dielectric slab placed in a uniform external field E₀ parallel to two of its faces. Every volume element Δv of the slab has a dipole moment PΔv in the direction of the field.

Anywhere inside the dielectric, the volume element Δv has no net charge (though it has net dipole moment). This is, because, the positive charge of one dipole sits close to the negative charge of the adjacent dipole.

At the surfaces of the dielectric normal to the electric field, there is a net charge density, the +ve ends of the dipoles remain unneutralised at the right surface and the –ve ends at the left surface. The unbalanced charges are the induced charges due to the external field. Thus the polarised dielectric is equivalent to two charged surfaces with induced surface charge densities, say σ_p and –σ_p.

The field produced by these surface charges opposes the external field. The total field in the dielectric is, thereby, reduced from the case when no dielectric is present. We should note that the surface charge density ± σ_p arises from bound (not free) charges in the dielectric.

Capacitors and Capacitance

A capacitor is a system of two conductors separated by an insulator.

• The conductors have charges; say q₁ and q₂, and potentials V₁ and V₂.


• Usually, in practice, the two conductors have charges q and – q, with potential difference V = V₁ – V₂ between them.

  The conductors may be so charged by connecting them to the two terminals of a battery. q is called the charge of the capacitor, though this, in fact, is the charge on one of the conductors – the total charge of the capacitor is zero.



• The electric field in the region between the conductors is proportional to the charge q. That is, if the charge on the capacitor is, say doubled, the electric field will also be doubled at every point.


• Potential difference V is the work done per unit positive charge in taking a small test charge from the conductor 2 to 1 against the field.


• V is proportional to q, and the ratio ${\displaystyle \frac{\mathrm{q}}{\mathrm{V}}}$ is a constant, ie. C = ${\displaystyle \frac{\mathrm{q}}{\mathrm{V}}}$. The constant C is called the capacitance of the capacitor.


• C is independent of q or V and depends only on the geometrical configuration (shape, size, separation) of the system of two conductors.


• The SI unit of capacitance is 1 farad (=1 coulomb volt^-1) or 1 F = 1 C V^–1.

  Other units: 1 μF = 10^–6 F, 1 nF = 10^–9 F, 1 pF = 10^–12 F



• A capacitor with fixed capacitance is symbolically shown as

  

  and a variable capacitor by

  



• A capacitor with large capacitance can hold large amount of charge q at a relatively small V.


• The maximum electric field that a dielectric medium can withstand without break-down (of its insulating property) is called its dielectric strength; for air it is about 3 × 10⁶ Vm^–1.


• There is a limit to the amount of charge that can be stored on a given capacitor without significant leaking.

Parallel Plate Capacitor Without Dielectric

A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance.

Let A be the area of each plate and d the separation between them. The two plates have charges Q and –Q. Since d is much smaller than the linear dimension of the plates (d² << A), we can use the result on electric field by an infinite plane sheet of uniform surface charge density.

Plate 1 has surface charge density σ = ${\displaystyle \frac{\mathrm{Q}}{\mathrm{A}}}$ and plate 2 has a surface charge density –σ. The electric field in different regions is:

Outer region I (region above the plate 1),

$$\mathrm{E}\mathrm{ }=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_{\mathrm{o}}}\mathrm{ }-\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_{\mathrm{o}}}\mathrm{ }=\mathrm{ }0$$

Outer region II (region belo the plate 2),

$$\mathrm{E}\mathrm{ }=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_{\mathrm{o}}}\mathrm{ }-\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_{\mathrm{o}}}\mathrm{ }=\mathrm{ }0$$

In the inner region between the plates 1 and 2, the electric fields due to the two charged plates add up, giving

$$\mathrm{E}\mathrm{ }=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_{\mathrm{o}}}+\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_{\mathrm{o}}}\mathrm{ }=\frac{\mathrm{\sigma }}{{\mathrm{\epsilon }}_{\mathrm{o}}}$$

$$=\frac{\mathrm{\sigma }\mathrm{A}}{{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{A}}$$

$$\Rightarrow \mathrm{E}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{A}}\mathrm{ }$$

The direction of electric field is from the positive to the negative plate.

Thus, the electric field is localized between the two plates and is uniform throughout. For plates with finite area, this will not be true near the outer boundaries of the plates.

The field lines bend outward at the edges – an effect called ‘fringing of the field’.

Since electric field is uniform, potential difference between the plates is,

$$\mathrm{V}\mathrm{ }=\mathrm{ }\mathrm{E}\mathrm{d}\mathrm{ }=\frac{\mathrm{q}\mathrm{d}}{{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{A}}$$

The capacitance C of the parallel plate capacitor is,

$$\mathrm{C}=\frac{\mathrm{q}}{\mathrm{V}}\mathrm{ }=\frac{{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{A}}{\mathrm{d}}$$

which depends only on the geometry of the capacitor and not on V or q.

Parallel Plate Capacitor With Dielectric Inserted

We have two large plates, each of area A, separated by a distance d. The charge on the plates is ± q, corresponding to the charge density ± σ (with σ = ${\displaystyle \frac{\mathrm{q}}{\mathrm{A}}}$).

Case I

When there is vacuum between the plates

$${\mathrm{E}}_{\mathrm{o}}\mathrm{ }=\frac{{\mathrm{\sigma }}_{\mathrm{o}}}{{\mathrm{\epsilon }}_{\mathrm{o}}}$$

And the potential difference V_o = E_o d

Therefore the capacitance,

$${\mathrm{C}}_{\mathrm{o}}=\mathrm{ }\frac{{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{A}}{\mathrm{d}}$$

Case II

Consider next a dielectric inserted between the plates fully occupying the intervening region.

The dielectric is polarised by the field, the effect is equivalent to two charged sheets (at the surfaces of the dielectric normal to the field) with surface charge densities σ_p and –σ_p and net surface charge density on the plates is ± (σ – σ_p ).

That is,

$$\mathrm{E}\mathrm{ }=\frac{\mathrm{\sigma }\mathrm{ }–\mathrm{ }{\mathrm{\sigma }}_{\mathrm{p}}}{{\mathrm{\epsilon }}_{\mathrm{o}}}$$

And

$$\mathrm{V}\mathrm{ }=\mathrm{ }\mathrm{E}\mathrm{ }\mathrm{d}\mathrm{ }=\frac{\mathrm{\sigma }\mathrm{ }–\mathrm{ }{\mathrm{\sigma }}_{\mathrm{p}}}{{\mathrm{\epsilon }}_{\mathrm{o}}}\mathrm{d}$$

For linear dielectrics, σ_p is proportional to E_o, i.e., to σ. Thus, (σ – σ_p) is proportional to σ,

$$\mathrm{\sigma }\mathrm{ }–\mathrm{ }{\mathrm{\sigma }}_{\mathrm{p}}=\frac{\mathrm{\sigma }}{\mathrm{K}}$$

where K is a constant characteristic of the dielectric.

Clearly, K >1.

That is,

$$\mathrm{V}\mathrm{ }=\mathrm{ }\frac{\mathrm{\sigma }\mathrm{ }}{{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{K}}\mathrm{d}=\mathrm{ }\frac{\mathrm{q}\mathrm{ }}{\mathrm{A}{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{K}}\mathrm{d}\mathrm{ }\mathrm{ }$$

And

$$\mathrm{C}=\frac{{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{K}\mathrm{ }\mathrm{A}}{\mathrm{d}}$$

ε = ε_o K is called the permittivity of the medium.

Relative Permittivity/Dielectric Constant

The dimensionless ratio K = ${\displaystyle \frac{\mathrm{\epsilon }}{{\mathrm{\epsilon }}_{\mathrm{o}}}}$ is called the dielectric constant or relative permittivity of the substance.

Also K = ${\displaystyle \frac{\mathrm{C}}{{\mathrm{C}}_{\mathrm{o}}}}$ K

The dielectric constant of a substance is the factor (>1) by which the capacitance increases from its vacuum value, when the dielectric is inserted fully between the plates of a capacitor.

Capacitors in series

When +ve plate of one capacitor is connected to the negative plate of another, the capacitors are said to be in series. In the series combination, charges on the two plates (± q) are the same on each capacitor. The total potential drop V across the combination is the sum of the potential drops V₁ and V₂ across C₁ and C₂, respectively.

$$\mathrm{V}\mathrm{ }=\mathrm{ }{\mathrm{V}}_1\mathrm{ }+\mathrm{ }{\mathrm{V}}_2\mathrm{ }=\frac{\mathrm{q}}{{\mathrm{C}}_1}+\frac{\mathrm{q}}{{\mathrm{C}}_2}\mathrm{ }\mathrm{ }\mathrm{ }$$

We can regard the combination as an effective capacitor with charge q and potential difference V.

The effective capacitance of the combination is,

C = ${\displaystyle \frac{\mathrm{q}}{\mathrm{V}}}$

⇒ V = ${\displaystyle \frac{\mathrm{q}}{\mathrm{C}}}$.

That is

$$\frac{\mathrm{V}}{\mathrm{Q}}=\frac{1}{\mathrm{C}}=\frac{1}{{\mathrm{C}}_1}+\frac{1}{{\mathrm{C}}_2}$$

Capacitors in parallel

In this case, the same potential difference is applied across both the capacitors. But the plate charges (± q₁) on capacitor 1 and the plate charges (±q₂) on the capacitor 2 are not necessarily the same:

q₁ = C₁V, q₂ = C₂V.

The equivalent capacitor is one with charge q = q₁ + q₂ and potential difference V. ie.

q = CV = C₁V + C₂V

Or the effective capacitance is,

C = C₁ + C₂

If the +ve plates of both the capacitors are connected to one point and the –ve plates to another, then

$$\mathrm{V}\mathrm{ }=\frac{{\mathrm{C}}_1{\mathrm{V}}_1+{\mathrm{C}}_2{\mathrm{V}}_2}{{\mathrm{C}}_1+{\mathrm{C}}_2}$$

If the +ve plates of one is connected to –ve plate of another, then

$$\mathrm{V}\mathrm{ }=\frac{{\mathrm{C}}_1{\mathrm{V}}_1-{\mathrm{C}}_2{\mathrm{V}}_2}{{\mathrm{C}}_1-{\mathrm{C}}_2}$$

Energy Stored in a Capacitor

Consider initially two uncharged conductors 1 and 2. Imagine next a process of transferring +ve charge from conductor 2 to conductor 1 bit by bit, so that at the end, conductor 1 gets charge q. By charge conservation, conductor 2 has charge –q at the end.

In transferring +ve charge from conductor 2 to conductor 1, work will be done externally, since at any stage conductor 1 is at a higher potential than conductor 2.

To calculate the total work done, we first calculate the work done in a small step involving transfer of an infinitesimal amount of charge. Consider the intermediate situation when the conductors 1 and 2 have charges q' and –q' respectively. At this stage, the potential difference V^' between conductors 1 to 2 are q'/C, where C is the capacitance of the system.

Next, a small charge δq' is transferred from conductor 2 to 1. Work done in this step (δW'), resulting in charge q' on conductor 1 increasing to q'+ δq', is given by

$$\mathrm{W}=\mathrm{V}\mathrm{\text{'}}\mathrm{q}\mathrm{\text{'}}=\frac{\mathrm{q}\mathrm{\text{'}}\mathrm{q}\mathrm{\text{'}}}{\mathrm{C}}$$

We can write,

$$\mathrm{ }\mathrm{q}\mathrm{\text{'}}\mathrm{q}\mathrm{\text{'}}\frac{\mathrm{ }1}{2}\left[{\left(\mathrm{q}\mathrm{\text{'}}+\mathrm{q}\mathrm{\text{'}}\right)}^2-{\mathrm{q}\mathrm{\text{'}}}^2\right]$$

Therefore,

$$\mathrm{W}=\frac{1}{2\mathrm{C}}\left[{\left(\mathrm{q}\mathrm{\text{'}}+\mathrm{q}\mathrm{\text{'}}\right)}^2-{\mathrm{q}\mathrm{\text{'}}}^2\right]$$

Both the equations are identical because the term of second order in δq′, i.e., δq′²/2C, is negligible, since δq′ is arbitrarily small.

The total work done (W) is the sum of the small work (δW) over the very large number of steps involved in building the charge q^' from zero to q.

$$\mathrm{W}\mathrm{ }=\sum _{\begin{array}{c}\mathrm{S}\mathrm{u}\mathrm{m}\mathrm{ }\mathrm{o}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{ }\\ \mathrm{a}\mathrm{l}\mathrm{l}\mathrm{ }\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{p}\mathrm{s}\end{array}}\mathrm{\delta }\mathrm{W}$$

$$=\sum _{\begin{array}{c}\mathrm{S}\mathrm{u}\mathrm{m}\mathrm{ }\mathrm{o}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{ }\\ \mathrm{a}\mathrm{l}\mathrm{l}\mathrm{ }\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{p}\mathrm{s}\end{array}}\frac{1}{2\mathrm{C}}\left[{\left(\mathrm{q}\mathrm{\text{'}}+\mathrm{q}\mathrm{\text{'}}\right)}^2-{\mathrm{q}\mathrm{\text{'}}}^2\right]$$

$$=\mathrm{ }\frac{1}{2\mathrm{C}}\left[\left\{{(\mathrm{q}\mathrm{\text{'}}}^2-0)\right\}+\left\{{\left(2\mathrm{q}\mathrm{\text{'}}\right)}^2-({\mathrm{q}\mathrm{\text{'}})}^2\right\}+\left\{{\left(3{\mathrm{q}}^{\mathrm{\text{'}}}\right)}^2-(2{\mathrm{q}\mathrm{\text{'}})}^2\right\}\mathrm{ }\dots ..\mathrm{ }+\mathrm{ }\left\{{\mathrm{q}}^2-(\mathrm{q}-\mathrm{q}\mathrm{\text{'}}{)}^2\right\}\right]$$

$$=\mathrm{ }\frac{{\mathrm{q}}^2}{2\mathrm{C}}$$

The same result can be obtained directly by integration

$$\mathrm{W}=\mathrm{ }{\int }_0^{\mathrm{q}}\frac{\mathrm{q}\mathrm{\text{'}}}{\mathrm{C}}\mathrm{ }\mathrm{d}\mathrm{q}\mathrm{\text{'}}=\mathrm{ }{\left[\mathrm{ }\frac{{\mathrm{q}\mathrm{\text{'}}}^2}{2\mathrm{C}}\mathrm{ }\right]}_0^{\mathrm{q}}=\mathrm{ }\frac{{\mathrm{q}}^2}{2\mathrm{C}}$$

Or Finally,

$$\mathrm{W}=\mathrm{ }\frac{1}{2}\mathrm{C}{\mathrm{V}}^2=\frac{1}{2}\mathrm{q}\mathrm{ }\mathrm{V}=\frac{1}{2}\frac{{\mathrm{q}}^2}{\mathrm{C}}=\frac{{\left(\mathrm{A}\right)}^2}{2}\times \frac{\mathrm{d}}{{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{A}}$$

This is the energy stored in a capacitor.

Since

$$\mathrm{E}\mathrm{ }=\frac{\mathrm{\sigma }}{{\mathrm{\epsilon }}_{\mathrm{o}}}$$

We can write

$$\mathrm{U}\mathrm{ }=\mathrm{ }\frac{1}{2}{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{ }{\mathrm{E}}^2\mathrm{ }\times \mathrm{ }\mathrm{A}\mathrm{ }\mathrm{d}=\mathrm{ }\frac{1}{2}{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{ }{\mathrm{E}}^2\mathrm{ }\mathrm{v}$$

Here, A d = v is the volume of the region between the plates (where alone the electric field exists).

If we define energy density u, as energy stored per unit volume of space,

$$\mathrm{u}=\mathrm{ }\frac{1}{2}{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{ }{\mathrm{E}}^2$$

Spherical Capacitor

The capacitance for spherical or cylindrical conductors can be obtained by evaluating the voltage difference between the conductors for a given charge on each.

For a charged conducting sphere, by applying Gauss' law, the electric field outside it is,

$$\mathrm{E}=\frac{\mathrm{q}}{4\mathrm{\pi }{\epsilon }_{\mathrm{o}}{\mathrm{r}}^2}$$

The voltage between the spheres can be found by integrating the electric field along a radial line:

$$\mathrm{\Delta }\mathrm{V}=\frac{\mathrm{q}}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}}{\int }_{\mathrm{a}}^{\mathrm{b}}\frac{1}{{\mathrm{r}}^2}\mathrm{ }\mathrm{d}\mathrm{r}\mathrm{ }$$

$$=\mathrm{ }\frac{\mathrm{q}}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}}\left[\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}\right]$$

From the definition of capacitance,

$$\mathrm{C}\mathrm{ }=\frac{\mathrm{q}}{\mathrm{\Delta }\mathrm{V}}=\frac{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}}{\left[\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}\right]}$$

• For a solid sphere, b = infinity and a = R, hence

  C = 4πε_oR

Cylindrical Capacitor

For a cylindrical geometry like a coaxial cable, the capacitance is usually stated as a capacitance per unit length. The charge resides on the outer surface of the inner conductor and the inner wall of the outer conductor.

The capacitance expression is

$$\frac{\mathrm{C}}{\mathrm{L}}=\frac{\mathrm{q}}{\mathrm{\Delta }\mathrm{V}}=\frac{2\mathrm{\pi }\mathrm{ }\mathrm{K}{\mathrm{\epsilon }}_{\mathrm{o}}}{\mathrm{l}\mathrm{n}\left[\frac{\mathrm{b}}{\mathrm{a}}\right]}$$

Van de Graaf Generator

Defintion of Van de Graaf Generator

It is a machine capable of building up potential difference of a few million volts, and fields close to the breakdown field of air which is about 3 × 10⁶ V/m.

Structure of Van de Graaf Generator

A Van de Graaff generator consists of a large spherical conducting shell (a few metre in diameter). By means of a moving belt and suitable brushes, charge is continuously transferred to the shell and potential difference of the order of several million volts is built up, which can be used for accelerating charged particles.

Principle of Van de Graaf Generator

Suppose we have a large spherical conducting shell of radius R, on which we place a charge Q. This charge spreads itself uniformly all over the sphere.

The field outside the sphere is just that of a point charge Q at the centre; while the field inside the sphere vanishes.

So the potential outside is that of a point charge; and inside it is constant, namely the value at the radius R. We thus have:

Potential inside conducting spherical shell of radius R carrying charge Q = constant

$$=\frac{\mathrm{Q}}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{R}}$$

Now, let us suppose that in some way we introduce a small sphere of radius r, carrying some charge q, into the large one, and place it at the centre.

Potential due to small sphere of radius r carrying charge q, at the surface of small sphere = ${\displaystyle \frac{\mathrm{q}}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{r}}\mathrm{ }}$

Potential due to small sphere of radius R carrying charge Q, at the surface of bigger sphere = ${\displaystyle \frac{\mathrm{Q}}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{R}}}$

Taking both charges q and Q into account we have the total potential, at the surface of smaller sphere,

$$\mathrm{V}\left(\mathrm{r}\right)=\mathrm{ }\frac{1}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}}\left(\frac{\mathrm{Q}}{\mathrm{R}}+\frac{\mathrm{q}}{\mathrm{r}}\right)$$

And, at the surface of the bigger sphere,

$$\mathrm{V}\left(\mathrm{R}\right)=\mathrm{ }\frac{1}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}}\left(\frac{\mathrm{Q}}{\mathrm{R}}+\frac{\mathrm{q}}{\mathrm{R}}\right)\mathrm{ }$$

$$\mathrm{V}\left(\mathrm{r}\right)-\mathrm{V}\left(\mathrm{R}\right)\mathrm{ }=\mathrm{ }\frac{\mathrm{q}}{4\mathrm{\pi }{\mathrm{\epsilon }}_{\mathrm{o}}}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$$

Assume now that q is positive. We see that, independent of the amount of charge Q that may have accumulated on the larger sphere and even if it is positive, the inner sphere is always at a higher potential: the difference V(r) –V(R) is positive. The potential due to Q is constant upto radius R (present in both terms) and so cancels out in the difference!

This means that if we now connect the smaller and larger sphere by a wire, the charge q on the former will immediately flow onto the matter, even though the charge Q may be quite large.

We can thus keep piling up larger and larger amount of charge on the latter. The potential at the outer sphere would also keep rising, at least until we reach the breakdown field of air.

This is the principle of the Van de Graaff generator.