Question

A nonconducting sphere has radius R = 2.81 cm and uniformly distributed charge q = +2.35 fC. Take the electric potential at the sphere's center to be V0 = 0. What is V at radial distance from the center (a) r = 1.60 cm and (b) r = R? (Hint: See an expression for the electric field.)

Answer 1

Answer:

(a). The electric potential at 1.650 cm is $-1.219\times10^{-4}\ V$.

(b). The electric potential at 2.81 cm is $-3.759\times10^{-4}\ V$.

Explanation:

Given that,

Radius of sphere R=2.81 cm

Charge = +2.35 fC

Potential at center of sphere

$V = 0$

(a). We need to calculate the potential at a distance r = 1.60 cm

Using formula of potential difference

$V_(r)-V_(0)=-\int_{0}^{r}{E(r)}dr$

$V_{r}-0=-\int_{0}^{r}{\dfrac{qr}{4\pi\epsilon_{0}R^3}}dr$

$V_{r}=-(\dfrac{qr^2}{8\pi\epsilon_{0}R^3})_{0}^{1.60\times10^{-2}}$

$V_{r}=-(\dfrac{2.35\times10^{-15}\times(1.60\times10^{-2})^2}{8\times\pi\times8.85\times10^{-12}\times(2.81\times10^{-2})^3})$

$V_{r}=-0.00012190\ V$

$V_{r}=-1.219\times10^{-4}\ V$

The electric potential at 1.650 cm is $-1.219\times10^{-4}\ V$.

(b). We need to calculate the potential at a distance r = R

Using formula of  potential difference

$V_{R}=-\dfrac{2.35\times10^{-15}}{8\pi\times8.85\times10^{-12}\times2.81\times10^{-2}}$

$V_{R}=-0.0003759\ V$

$V_{R}=-3.759\times10^{-4}\ V$

The electric potential at 2.81 cm is $-3.759\times10^{-4}\ V$.

Hence, This is the required solution.