Question

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?

Answer 1

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

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

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

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