Question

The equipotential surface for a point charge of 1nc has potential of 9v . find the radius of the equipotential surface?​

Answer 1

Answer:

W = Fd cos θ = qEd cos θ = 0.

Answer 2

Answer:

The radius of the surface is 1m.

Explanation:

This is the formula-based question.

Potential due to point charge is given as,

$V=\frac{kQ}{r}$             (1)

Where,

V=potential of the surface

k=columbs constant=9×10⁹Nm²C⁻²

Q=charge on the given  surface

Now the values given in the question are,

V=9v

Q=1 nC=1×10⁻⁹C

By substituting the required values in equation (1) we get;

$9=\frac{9\times 10^9\times 10^-^9}{r}$

$r=\frac{9\times 10^9\times 10^-^9}{9}=1m$

Hence, the radius of the equipotential surface is 1m.