16.2 Electric Potential

16.2 Electric Potential

In order to give a concept of electric potential at a point in an electric field due to a point charge, we must have a reference to which we assign zero electric potential. This point is usually taken at infinity.

If we take A to be at infinity and choose $V_{A} = 0$, the electric potential at B will be $V_{B} = W_{\infty B}/q_{o}$, or dropping the subscripts

which states that:

The electric potential at any point in an electric field is equal to work done in bringing a unit positive charge from infinity to that point keeping it in electrostatic equilibrium.

It is to be noted that potential at a point is still potential difference between the potential at that point and potential at infinity. Both potential and potential differences are scalar quantities because both $W$ and $q_{o}$ are scalars.

Let us derive an expression for the potential at a certain point in the field of a positive point charge $q$. This can be accomplished by bringing a unit positive charge from infinity to that point keeping the charge in equilibrium. The target can be achieved by using Eq. 16.5 in the form $\Delta V = -E \Delta r$, provided electric intensity $E$ remains constant. However in this case $E$ varies inversely as square of distance from the point charge, it no more remains constant so we use basic principles to compute the electric potential at a point. The field is radial as shown in Fig. 16.2.

(Fig. 16.2)

Let us take two points A and B, infinitesimally close to each other, so that $E$ remains almost constant between them. The distances of points A and B from $q$ are $r_{A}$ and $r_{B}$ respectively and distance of midpoint of space interval between A and B is $r$ from $q$. Then according to Fig. 16.2

As $r$ represents midpoint of interval between A and B so

The magnitude of electric intensity at this point is,

As the points A and B are very close then, as a first approximation, we can take the arithmetic mean to be equal to geometric mean which gives

Therefore, $r^{2} = r_{A} r_{B} \quad \ldots (16.11)$

Thus, Eq. 16.10 can be written as

Now, if a unit positive charge is moved from B to A, the work done is equal to the potential difference between A and B,

Substituting value of $E$ from Eq. 16.12.

To calculate absolute potential or potential at A, point B is assumed to be infinity point so that $V_{B} = 0$, and hence

This gives,

The general expression for electric potential $V$ at a distance $r$ from $q$ is,

The electric potential will be negative if the charge $q$ is negative. However, if there are more than one charge at different locations, then each charge contributes to the total electric potential. At any point, the total electric potential is the algebraic sum of the individual potentials created by all the charges.

Example 16.2

Find the electric potential at a point $1.2\,\mathrm{m}$ from a charge of $5.0 \times 10^{-8}\,\mathrm{C}$.

Solution

$q = 5.0 \times 10^{-8}\,\mathrm{C}$, $\quad r = 1.2\,\mathrm{m}$, $\quad V = ?$

Using $V = \dfrac{1}{4\pi\epsilon_{o}} \dfrac{q}{r}$

Putting the values $V = \dfrac{(9 \times 10^{9}\,\mathrm{N\,m^{2}\,C^{-2}})(5.0 \times 10^{-8}\,\mathrm{C})}{1.2\,\mathrm{m}}$

$V = 375\,\mathrm{V}$

Example 16.3

Two point charges $+8\,\mu\mathrm{C}$ and $-4\,\mu\mathrm{C}$ are placed $60\,\mathrm{cm}$ apart as shown in Fig. 16.3. Where is the electric potential zero on the line passing through the charges?

(Fig. 16.3)

Solution

The electric potential at any point will be zero where the potential due to positive charge will be exactly equal in magnitude to that due to negative charge. Since the magnitude of positive charge is greater than that of negative charge, so there will be two zero potential points. Both the points should be closer to the smaller negative charge. That is one point (A) between the two charges and the other point (B) outside the negative charge.

Let the point A lies at a distance $x$ from the positive charge. Then using we have,

Let the point A lies at a distance $x$ from the positive charge. Then using $V= \dfrac{1}{4\pi\epsilon_{o}} \dfrac{q}{r}$ we have,

Electric potential at A due to positive charge $= \dfrac{1}{4\pi\epsilon_{o}} \left(\dfrac{8 \times 10^{-6}\,\mathrm{C}}{x}\right)$

Electric potential at A due to negative charge $= \dfrac{1}{4\pi\epsilon_{o}} \left(\dfrac{4 \times 10^{-6}\,\mathrm{C}}{60\,\mathrm{cm} - x}\right)$

As both the potentials balance each other, therefore

$\dfrac{1}{4\pi\epsilon_{o}} \left(\dfrac{8 \times 10^{-6}\,\mathrm{C}}{x}\right) = \dfrac{1}{4\pi\epsilon_{o}} \left(\dfrac{4 \times 10^{-6}\,\mathrm{C}}{60\,\mathrm{cm} - x}\right)$

or $480\,\mathrm{cm} - 8x = 4x$

$12x = 480\,\mathrm{cm}$

$x = 40\,\mathrm{cm}$

Electric potential at B due to positive charge $= \dfrac{1}{4\pi\epsilon_{o}} \left(\dfrac{4 \times 10^{-6}\,\mathrm{C}}{60\,\mathrm{cm} - x}\right)$

Electric potential at B due to negative charge $= \dfrac{1}{4\pi\epsilon_{o}} \left(\dfrac{4 \times 10^{-6}\,\mathrm{C}}{y}\right)$

Applying balance condition,

$\dfrac{1}{4\pi\epsilon_{o}} \left(\dfrac{8 \times 10^{-6}\,\mathrm{C}}{x}\right) = \dfrac{1}{4\pi\epsilon_{o}} \left(\dfrac{4 \times 10^{-6}\,\mathrm{C}}{60\,\mathrm{cm} - x}\right)$

$4 (60\,\mathrm{cm} + y) = 8y$

$240\,\mathrm{cm} + 4y) = 8y$

$4y = 240\,\mathrm{cm}$

$y = \dfrac{240\,\mathrm{cm}}{4} = 60\,\mathrm{cm}$