Question

A solid sphere with a radius of 50.0 cm has a total positive charge of 40.0 μC uniformly distributed in its volume. Calculate the magnitude of the electric field 10.0 cm away from the center of the sphere:
a

1.8e+008 N/C

b

3.2e+004 N/C

c

1.44e+010 N/C

d

2.880e+005 N/C

Answer 1

Answer:

$E=2.88\times 10^5\ N/C$                              

Explanation:

It is given that,

The radius of the solid sphere, R = 50 cm = 0.5 m

Charge on the sphere, $q=40\ \mu C=40\times 10^{-6}\ C$

We need to find the magnitude of the electric field r = 10.0 cm away from the center of the sphere. The electric field at point r away form the center of the sphere is given by :

$E=\dfrac{kqr}{R^3}$

$E=\dfrac{9\times 10^9\times 40\times 10^{-6}\times 0.1}{0.5^3}$

$E=2.88\times 10^5\ N/C$

So, the electric field 10.0 cm away from the center of the sphere is $2.88\times 10^5\ N/C$. Hence, this is the required solution.