A hollow metal sphere has inner and outer radii of 20.0 cm and 30.0 cm, respecti
ID: 2148532 • Letter: A
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 = 6.93?104 N/C, directed radially inward. At point Q, a distance of 35.0 cm from the center, the electric field is found to be E2 = 6.93?104 N/C, directed radially outward. a) Determine the total charge on the surface of the inner sphere. b) Determine the total charge on the surface of the inner surface of the hollow sphere. c) Determine the total charge on the surface of the outer surface of the hollow sphere
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 = 6.93?104 N/C, directed radially inward. At point Q, a distance of 35.0 cm from the center, the electric field is found to be E2 = 6.93?104 N/C, directed radially outward. a) Determine the total charge on the surface of the inner sphere. b) Determine the total charge on the surface of the inner surface of the hollow sphere. c) Determine the total charge on the surface of the outer surface of the hollow sphereExplanation / Answer
Since E = kq/r2, q = Er2/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.
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