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zaharov [31]
3 years ago
5

A solid metal ball of radius 1.5 cm bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing a

uniformly distributed charge of -7 nC (see the figure) on its outer surface. The distance between the centers of the balls is 9 cm.Assume the +x axis runs to the right, the +y axis runs up and the +z axis is out.What is the electric field at the center of the metal ball due only to the charges on the surface of the metal ball?
Physics
1 answer:
blsea [12.9K]3 years ago
4 0

We have that the electric field at the center of the metal ball due only to the charges on the surface of the metal ball is

E=7*10^{9}N/C

From the question we are told that

A solid metal ball of radius 1.5 cm

bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing

uniformly distributed charge of -7 nC

The distance between the centers of the balls is 9 cm

Generally the equation for the electric field  is mathematically given as

E=\frac{kq_2}{d^2}\\\\E=\frac{(9*10^9)7*10^{-2}}{9*10^{-2}}\\\\

E=7*10^{9}N/C

For more information on this visit

brainly.com/question/21811998

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