Question

*The equipotential surface for a point charge of 1 nC has a potential of 9 V. The radius of the equipotential surface will be:* 1️⃣ 10 m 2️⃣ 20 m 3️⃣ 30 m 4️⃣ 1 m​

Answer 1

Given:

The equipotential surface for a point charge of 1 nC has a potential of 9 V.

To find:

Radius of equipotential surface?

Calculation:

• We know that fall of point charge the equipotential surfaces are always concentric spheres at different radii from the charge.

Let the radius for that particular equipotential surface be 'r'.

$V = \dfrac{1}{4 \pi \epsilon_{0}} \times \dfrac{q}{r}$

$\implies 9 = (9 \times {10}^{9} ) \times \dfrac{1 \times {10}^{ - 9} }{r}$

$\implies 9 = \dfrac{9 \times {10}^{(9 - 9)} }{r}$

$\implies 9 = \dfrac{9 \times {10}^{(0)} }{r}$

$\implies 9 = \dfrac{9}{r}$

$\implies \: r = 1 \: metre$

So, radius of equipotential surface is 1 metres.