Question

The electric field 14.0 cm from the surface of a copper ball of radius 2.0 cm is directed toward the ball's center and has magnitude 9.0 ✕ 10^2 N/C. How much charge is on the surface of the ball (in C)? (Include the sign of the value in your answer.)

Answer 1

Answer:

Q = - 256 X 10⁻⁷ C .

Explanation:

Electric field due to a charge Q at a distance d from the center is given by the expression

E = k Q /d² Where k is a constant and it is equal to 9 x 10⁹

Put the given value in the equation

9 x 10² = $\frac{9\times10^9\times Q}{(14+2)^2\times10^{-4}}$

Q = $\frac{9\times16^2\times10^{-2}}{9\times10^9}$

Q = - 256 X 10⁻⁷ C .

It will be negative in nature as the field is directed towards the center.