A solid metal ball of radius 1.5 cm bearing a charge of −17 nC is located near a solid plastic ball of radius 2 cm bearing a uniformly distributed charge of +7 nC (Figure 15.62) on its outer surface. The distance between the centers of the balls is 9 cm. (a) Show the approximate charge distribution in and on each ball. (b) What is the electric field at the center of the metal ball due only to the charges on the plastic ball? (c) What is the net electric field at the center of the metal ball? (d) What is the electric field at the center of the metal ball due only to the charges on the surface of the metal ball?

The given data can be listed below,

• The radius of solid metal ball is,$r_m=1.5\mathrm{cm}$
• The charge on solid metal ball is,$q_m=-17nC$
• The radius of the plastic ball is,localid="1656935508662" $r_p=2\mathrm{cm}$
• The charge on the plastic ball is,localid="1656935521508" $q_p=+17n\mathrm{C}$

The term "continuous quantity" refers to a quantity with no interruptions in between, i.e., a particle distribution that is uniform. The charge density is consistent across the distribution.

The positive plastic ball doesn't polarize the charges inside the metal ball as none exists. Still, it polarizes the charges on the surface by redistributing the negative charges where it attracts the negative charges. It is more concentrated on the side of the sphere closest to the positive plastic ball. The negatively charged metal ball can't polarize the charges on the surface of the plastic as they can't move freely. Still, instead, it polarizes the molecules of the plastic inside such it induces a dipole with positive charges being attracted or pointing toward the negative metal ball.it can be shown by figure below.

The electric field is given by,

$\mathrm{E}=\frac{\mathit{\text{Kq}}}{r^{\mathit{2}}}$

Here, K is the coulomb constant whose value is $9x{10}^9{\mathrm{Nm}}^{2}l{C}^2$, q is the charge on the sphere, and r is the distance of plastic ball to the center of metal ball.

Substitute all the values in the above,

$E\mathit{=}\frac{(9\times {10}^9{\mathrm{Nm}}^2/{\mathrm{C}}^2(17\times {10}^{-9}\mathrm{C})}{{(0.09\mathrm{m})}^2}$

$=18.88\times {10}^3\mathrm{N}/\mathrm{C}$

Thus, the electric field at center of metallic ball due to plastic ball is $=18.88\times {10}^3\mathrm{N}/\mathrm{C}$

The charges on the surface of the sphere are not uniformly distributed, i.e., are being polarized by an external field. The external source of the electric field would have a value inside the metallic sphere. The polarized charges are redistributed in a manner such that it has a value equal to the electric field of the external source and opposite in direction.

The net electric field is given by,

$E_{net}=E_{external}+E_{polarised}$

$=0$

Thus, the net electric field on uniformly distributed charged metal bond is zero.

The polarized electric field at the center of metal ball is given by,

$E_{polarised}=-E_{external}$

The value of external electric field is $18.88x{10}^3\mathrm{N}/\mathrm{C}$
$18.88x{10}^3\mathrm{N}/\mathrm{C}.$

localid="1656935746813" $E_{polarised}=-18.88x{10}^3\mathrm{N}/\mathrm{C}$

Thus, the magnitude of electric field at the center of metal ball due to polarized charge on the surface of ball is $18.88x{10}^3\mathrm{N}/\mathrm{C}$and its direction is towards plastic ball.