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motikmotik
3 years ago
10

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 magni

tude 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.)
Physics
1 answer:
Marina CMI [18]3 years ago
5 0

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.

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