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

Q = - 256 X 10⁻⁷ C .

Step-by-step 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 =
`(9*10^9* Q)/((14+2)^2*10^(-4))`

Q =
`(9*16^2*10^(-2))/(9*10^9)`

Q = - 256 X 10⁻⁷ C .

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