Question

A hollow metal sphere has inner and outer radii of 20.0 cm and 30.0 cm, respectively. As shown in the figure, a solid metal sphere of radius 10.0 cm is located at the center of the hollow sphere. The electric field at a point P, a distance of 15.0 cm from the center, is found to be E₁ = 5.53

Answer 1

Since E = kq/r², q = Er^2/k^. which for part (a) works out to (-5.53 * 10⁴ N/C)(0.15 m)²/(8.988 * 10⁹ Nm²/C²) = -1.384 * 10^-7 C. Note that since you are inside the hollow sphere whatever charge is on there is irrelevant. Also note the charge is negative since the E vector points in.

(b) is 1.384 * 10^-7 C , the charge on the inner surface of the hollow sphere will be equal and opposite to the charge in the center since the E field inside the hollow sphere has to be zero. Any Gaussian surface inside the hollow sphere must enclose zero net charge.

For (c), calculate Er²/k again = (5.53 * 10⁴ N/C)(0.35 m)²/(8.988 * 10⁹ Nm²/C²) = 7.537 * 10^-7 C

This is the net charge enclosed by the Gaussian surface at 35 cm; but since the charge on the central sphere and the charge on the inner surface of the hollow sphere are equal and opposite, this must also be the charge on the outer surface of the hollow sphere.