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

`E=2.88* 10^5\ N/C`

Step-by-step 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* 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=(kqr)/(R^3)`

`E=(9* 10^9* 40* 10^(-6)* 0.1)/(0.5^3)`

`E=2.88* 10^5\ N/C`

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