Question

draw a curve representing change in electric potential due to point charge with distance ?​

Answer 1

To draw:

A curve representing change in electric potential due to point charge with distance

Calculation:

The general expression of Electrostatic Field Intensity by a point charge with distance :

$\boxed{\therefore \: E = \dfrac{1}{4\pi \epsilon_{0} } \bigg( \dfrac{q}{ {r}^{2} } \bigg)}$

Now , we know :

$\therefore \: dV = - E \times dr$

Integrating on both sides:

$\displaystyle \implies \: \int_{V}^{0}dV = - \int_{r}^{ \infty} E \times dr$

$\displaystyle \implies \: \int_{V}^{0}dV = - \int_{r}^{ \infty} \frac{q}{4\pi \epsilon_{0} {r}^{2} } \times dr$

$\displaystyle \implies \: \int_{V}^{0}dV = - \dfrac{q}{4\pi \epsilon_{0}} \int_{r}^{ \infty} \frac{dr}{{r}^{2} }$

$\displaystyle \implies \: V = \dfrac{q}{4\pi \epsilon_{0}} \bigg( \frac{1}{r} \bigg)$

So, the required expression:

$\boxed{ \bold{ \: V = \dfrac{q}{4\pi \epsilon_{0}} \bigg( \frac{1}{r} \bigg)}}$

So, the graph will be hyperbolic in shape.

• In the attached graph , V is in Y axis and r is on X axis.