Two concentric spheres are shown in the figure. The inner sphere is a solid nonc
ID: 1629743 • Letter: T
Question
Two concentric spheres are shown in the figure. The inner sphere is a solid nonconductor and carries a charge of +5.00 mu C uniformly distributed over its outer surface. The outer sphere is a conducting shell that carries a net charge of -8.00 mu C. No other charges are present. The radii shown in the figure have the values R_1 = 10.0 cm, R_2 = 20.0 cm, and R_3 = 30.0 cm. (k = 1/4 pi epsilon_0 = 8.99 times 10^9 N middot m^2/C^2) (a) Find the total excess charge on the inner and outer surfaces of the conducting sphere. (b) Find the magnitude and direction of the electric field at the following distances r from the center of the inner sphere: (i) r = 9.5 cm, (ii) r = 15.0 cm. (iii) r = 27.0 cm, (iv) r = 35.0 cm.Explanation / Answer
Given,
Net charge on inner sphere is 5 x 10-6 C
Net charge on outer sphere is 8 x 10-6 C
a) So total excess charge on the inner sphere will be -5uC, in order to neutralize the +5uC charge from the inner sphere's surface.
Similarly, total excess charge on the outer surface will be -3uC, in order to make the net charge equal to -8uC.
b) I) Electric field at distance r = 9.5 cm will be 0, because 9.5cm is not outside the inner nonconducting sphere.
II) Electric field at distance r = 15 cm will be 2 x 106 N/C
Because, E = kq/r2 = ((9 x 109) (5 x 10-6)) /(0.15)2
Direction would be radially outward, because E fields flow from higher potential to lower potential, i.e. +5uC to -5uC.
III) Electric field at distance r = 27 cm will be 0, because it is within the outer ring; no charge on inside, only surfaces
IV) Electric field at distance r = 35 cm will be 2.20 x 105 N/C
because E = kq/r2 = ((9 x 109)x(3 x 10-6)/(0.35)2)
Direction would be radially inward, same reason, but it is negative, which flows towards center.
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