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

Answer:

a

  The  value at an inside point is Zero  

b

The electric field is $E = 2.7*10^{6} \ N/C$

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

$E = \frac{kq}{ a^2}$

Here a is the position outside the spherical ball that is been considered and the value is $a = 10 \ cm = \frac{10}{100} = 0.1 \ m$

and k is the coulombs constant with value

$k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

=> $E = \frac{ 3 *10^{-6} * 9*10^9 }{ (0.1)^2}$

=> $E = 2.7*10^{6} \ N/C$