Question

When a charge is placed on a metal sphere, it ends up in equilibrium at the outer surface. Use this information to determine the electric field of +3.0 µC charge put on a 5.0-cm aluminum spherical ball at the following two points in space:

a. A point 1.0 cm from the center of the ball (an inside point).
b. A point 10 cm from the center of the ball (an outside point)

Answer 1

a

The value at an inside point is Zero

b

The electric field is
`E = 2.7*10^(6) \ N/C`

Step-by-step explanation:

From the question we are told that

The magnitude of the charge is
`q = 3.0 \mu C = 3.0 *10^(-6) \ C`

The radius of the spherical ball is
`r = 5.0 \ mm = 0.005 \ m`

Generally according to law postulated by Gauss the magnitude of charge enclosed inside a conducting material is zero which implies that the electric field inside the spherical ball is zero

Generally the electric field out side the spherical ball is mathematically represented as

`E = (kq)/( a^2)`

Here a is the position outside the spherical ball that is been considered and the value is
`a = 10 \ cm = (10)/(100) = 0.1 \ m`

and k is the coulombs constant with value

`k = 9*10^(9)\ kg\cdot m^3\cdot s^(-4) \cdot A^(-2).`

=>
`E = ( 3 *10^(-6) * 9*10^9 )/( (0.1)^2)`

=>
`E = 2.7*10^(6) \ N/C`