Question

231 Derive the expression for electric potential
out a point due to a point charge.​

Answer 1

Step-by-step explanation:

The ‘electric potential’ due to a point charge is kq/r.

Solution:

Let us consider q’ be the point charge placed at a distance r from the charge q.

By coulomb law, the force exerted on the point charge due to the positive charge q is

F=k \frac{q q^{\prime}}{r^{2}} \rightarrow(1)F=k

r

2

qq

′

→(1)

The electric potential is defined as the work done to bring a point charge to a particular position.

V=\frac{W}{q^{\prime}} \rightarrow(2)V=

q

′

W

→(2)

The work done can be determined by

W=-\int_{\infty}^{r} F . d r=-\int_{\infty}^{r} k \frac{q q^{\prime}}{r^{2}} d rW=−∫

∞

r

F.dr=−∫

∞

r

k

r

2

qq

′

dr

W=-k q q^{\prime}\left[-\frac{1}{r}\right]_{\infty}^{r}=\frac{k q q^{\prime}}{r}W=−kqq

′

[−

r

1

]

∞

r

=

r

kqq

′

Thus the ‘electric potential’ will be

V=\frac{\frac{k q q^{\prime}}{r}}{q^{\prime}}=\frac{k q}{r}V=

q

′

r

kqq

′

=

r

kq

The electric potential for a point charge is kq/r.