Question

A non conducting sphere of radius 0.04 m has a charge of 5.0 × 10^-9 C deposited on it. CalculateThe magnitude of the electric field at 0.02m from the center of the sphere

Answer 1

Answer:

The electric field at a distance r = 0.02 m is 14062.5 N/C.

Solution:

Refer to fig 1.

As per the question:

Radius of sphere, R = 0.04 m

Charge, Q = $5.0\times 10^{- 9} C$

Distance from the center at which electric field is to be calculated, r = 0.02 m

Now,

According to Gauss' law:

$E.dx = \frac{Q_{enclosed}}{\epsilon_{o}}$

Now, the charge enclosed at a distance r is given by volume charge density:

$\rho = \frac{Q_{enclosed}}{area}$

$\rho = \frac{Q_{enclosed}}{\frac{4}{3}\pi R^{3}}$

Also, the charge enclosed Q' at a distance r is given by volume charge density:

$\rho = \frac{Q'_{enclosed}}{\frac{4}{3}\pi r^{3}}$

Since, the sphere is no-conducting, Volume charge density will be constant:

Thus

$\frac{Q_{enclosed}}{\frac{4}{3}\pi R^{3}} = \frac{Q'_{enclosed}}{\frac{4}{3}\pi r^{3}}$

Thus charge enclosed at r:

$Q'_{enclosed} = \frac{Q_{enclosed}}{\frac{r^{3}}{R^{3}}$

Now, By using Gauss' Law, Electric field at r is given by:

$4\pi r^{2}E = \frac{Q_{enclosed}r^{3}}{\epsilon_{o}R^{3}}$

Thus

$E = \frac{Q_{enclosed}r}{4\pi\epsilon_{o}R^{3}}$

$E = \frac{(9\times 10^{9})\times 5.0\times 10^{- 9}\times 0.02}{0.04^{3}}$

E = 14062.5 N/C