Derive an expression for electric potential due to a point charge - Param Himalaya

Derive an expression for electric potential due to a point charge - Param Himalaya

2.3. ELECTRIC POTENTIAL DUE TO A POINT CHARGE

Derive an expression for electric potential due to a point charge. What is the nature of symmetry of electric potential due to single charge and why?

Consider a point charge q. Let a unit positive charge (+1) be placed at a point P in the electric field of the charge q. Let the distance between +q and unit positive charge (+1) be x.

ELECTRIC POTENTIAL DUE TO A POINT CHARGE

The electrostatic or Coulomb's force of repulsion between charge $q$ and +1 charge is given by

$F = \frac{1}{4\pi\epsilon_0} \frac{q \times 1}{x^2}$

or

$F = \frac{1}{4\pi\epsilon_0} \frac{q}{x^2}$...(1)

This force is directed radially outwards along the line joining +q charge and +1 charges (i.e., in the direction of $\vec{E}$) 

Now, let unit positive charge be displaced towards +q charge, without acceleration, such that the displacement of unit positive charge is $dx$.

Work done in displacing unit positive charge through $dx$ is given by

$dW = \vec{F} \cdot d\vec{x}$

$dW= F dx \cos 180^\circ = -F dx$ (\because \cos 180^\circ = -1)$

Using eqn. (1), we get,

$dW = -\frac{q}{4\pi\epsilon_0 x^2} dx$...(2)

Now, let the unit positive charge be displaced from $\infty$ to point A i.e., from $x = \infty$ to $x = r$. The total work done to do so can be calculated by integrating eqn. (2) between the limits $x = \infty$ to $x = r$.

$\int dW = \int_{x=\infty}^{x=r} -\frac{q}{4\pi\epsilon_0 x^2} dx$

$W = -\frac{q}{4\pi\epsilon_0} \int_\infty^r x^{-2} dx = -\frac{q}{4\pi\epsilon_0} \left[ \frac{x^{-1}}{-1} \right]_\infty^r$

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

$\left( \because \frac{1}{\infty} = 0 \right)$

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

But, work done in bringing a unit positive charge from infinity to a given point in the electric field of source charge = Electric potential due to the source charge at that point.

W = V(r)Electric potential at a distance r from the source charge +q)

$V(r) = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$ ...(3)

Thus, electric potential at a distance $r$ from a point charge $q$ is

(i) directly proportional to the charge $q$, and

(ii) inversely proportional to the distance $r$.

If point source charge $q$ is positive, then electric potential $V(r)$ is positive. However, if point source charge $q$ is negative, then electric potential $V(r)$ is negative. The value of $V(r)$ is large if $q$ is large and value of $V(r)$ is small if $q$ is small.

Variation of electric potential due to a charge q with distance r from the charge:

The variation of electric potential V with distance r from the charge 

for charge $q > 0$ (i.e., positive charge) and charge $q < 0$ (i.e., negative charge).

The variation of electric potential V with distance r from the charge

The variation of electric potential V due to a point charge q (positive or negative) with $\frac{1}{r}$, where r is the distance from the point charge q.

variation of electric potential V due to a point charge q (

Electric potential due to a single charge is spherical symmetric as the value of electric potential due to this charge is same at all points distant r around this charge.

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