# Electric Potential

###### Electric Potential

Electric Potential at a point in the electric field is defined as the work done in moving (without any acceleration) a unit positive charge from infinity to that point against the electrostatic force irrespective of the path followed.

Consider a point charge +q at O  & OP = r

Electric field strength at A is

$\large E = \frac{1}{4\pi \epsilon_0} \frac{q}{x^2}$ (here , OA = x)

Small amount of work done in moving a unit +ve charge from A to B is

$\large dW = \vec{E}.\vec{dx} = E dx cos180^o = – E dx$

Total amount of work done in moving a unit +ve charge from ∞ to point P is

$\large W = – \int_{\infty}^{r} E dx = – \frac{q}{4\pi \epsilon_0}\int_{\infty}^{r} \frac{1}{x^2}dx$

$\large W = – \frac{q}{4\pi \epsilon_0} [-\frac{1}{x}]_{\infty}^{r}$

$\large W = \frac{q}{4\pi \epsilon_0} [\frac{1}{r} – \frac{1}{\infty}]$

$\large W = \frac{1}{4\pi \epsilon_0} \frac{q}{r}$

By definition ;

Potential at P due to q is

$\large V = \frac{1}{4\pi \epsilon_0} \frac{q}{r}$

At r = ∞  , V = 0

NOTE : Electric potential difference between any two points (B and A) in an electric field is the negative of work done by electrical force per unit charge in moving a charge from A to B.

$\displaystyle V_B – V_A = -\frac{W_e}{q}$

$\displaystyle = \frac{U_B – U_A}{q}$ , (Since − We = ΔU)

In other words

$\large dV = -\vec{E}.\vec{dr}$

$\large \int_{V_A}^{V_B} dV = -\int_{r_A}^{r_B}\vec{E}.\vec{dr}$

$\large V_B – V_A = -\int_{r_A}^{r_B}\vec{E}.\vec{dr}$

For the electric field due to a point charge Q

$\large E = \frac{k Q}{r^2}$

$\displaystyle V_B – V_A = kQ ( \frac{1}{r_B}-\frac{1}{r_A} )$

Potential of a point at infinity is taken to be zero.

i.e. if rB = ∞ , VB = 0.

Now the potential of any point A is given as.

$\displaystyle V_A = \frac{kQ}{r_A}$

Potential is a scalar quantity.

Thus, the potential at a point due to more than one charge can be found simply by adding the potentials due each charge separately.

Potential at a point due to group of charges :

$\large V = V_1 + V_2 + V_3 + ….$

$\large V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r_1} + \frac{q_2}{r_2} + \frac{q_3}{r_3} + …..]$

Electric potential due to a positive charge is taken to be positive and that due to a negative charge is taken to be negative.

##### Potential due to a uniformly charged disc

The elemental ring’s area = 2πxdx

Let σ = Surface charge density

$\displaystyle dV = \frac{1}{4 \pi \epsilon_0} \frac{\sigma (2\pi x dx)}{\sqrt{r^2 + x^2}}$

$\displaystyle V = \int dV = \frac{\sigma}{2\epsilon_0 }\int_{0}^{R} \frac{x dx}{\sqrt{r^2+x^2}}$

$\displaystyle V = \frac{\sigma}{2\epsilon_0 } [\sqrt{R^2 + r^2 }- r ]$

##### Potential due to a uniformly charged spherical shell

If the charge on the shell = q

Case (i) When , r > R

$\displaystyle E = k \frac{q}{r^2}$

$\displaystyle V = -\int_{\infty}^{r} E dr$

$\displaystyle V = -\int_{\infty}^{r} \frac{k q}{r^2} dr$

$\displaystyle V = \frac{k q}{r}$

Case (ii) for r = R

$\displaystyle V = \frac{k q}{R}$

Case (iii) for r < R

$\displaystyle E = k \frac{q}{r^2}$ ; (R < r < ∞)

E = 0 , (r < R )

$\displaystyle V = -[\int_{\infty}^{R} E dr + \int_{R}^{r} E dr]$

$\displaystyle V = -\int_{\infty}^{R} \frac{k q}{r^2} dr – \int_{R}^{r} 0 dr$

$\displaystyle V = \frac{k q}{R}$

Exercise: An annular disc of inner radius a and outer radius ‘ 2a ‘ is uniformly charged with uniform surface charge density σ . Find the potential at a distance ‘ a ‘ from the centre at a point ‘ P ‘ lying on the axis.

Exercise : Calculate the potential due to a thin charged rod of length L at the points along and perpendicular to the length. Charge per unit length on the rod is λ.