ELECTRIC POTENTIAL AND CAPACITANCE

Ans.

Consider a point charge $+q$ at the origin as shown in the figure. Work done for $dx$ is $dW$.$$\therefore dW=-Fdx\cos\theta$$$$dW=-Fdx\quad ( \because \cos\theta =1)$$
The total work done in bringing a test positive charge $q_{\circ}$ from infinity to the point $P$ $$W=-\:\int_{\infty}^{r} \frac{1}{4\pi\varepsilon_{\circ}}\frac{qq_{\circ}}{x^{2}}dx$$

$$W=-\:\frac{qq_{\circ}}{4\pi\varepsilon_{\circ}}\int_{\infty}^{r} \frac{1}{x^{2}}dx$$

$$W=-\:\frac{qq_{\circ}}{4\pi\varepsilon_{\circ}} \left [ \frac{-1}{x} \right ]_{\infty}^{r} $$

$$W=-\:\frac{qq_{\circ}}{4\pi\varepsilon_{\circ}} \left [ \frac{-1}{r} +\frac{1}{\infty} \right ] $$

$$W= \frac{1}{4\pi\varepsilon_{\circ}} \frac{qq_{\circ}}{r} $$

$$\frac{W}{q_{\circ}}= \frac{1}{4\pi\varepsilon_{\circ}} \frac{q}{r} $$$$\therefore \text{Electric Potential}\:\: V = \frac{1}{4\pi\varepsilon_{\circ}} \frac{q}{r} $$