For the configuration shown in the figure below, suppose a = 5.00 cm, b = 20.0 c

Question

For the configuration shown in the figure below, suppose a = 5.00 cm, b = 20.0 cm, and c = 25.0 cm. Furthermore, suppose the electric field at a point 18.0 cm from the center is measured to be 3. 15 times 10^3 N/C radially inward and the electric field at a point 50.0 cm from the center is of magnitude 235 N/C and points radially outward. From information, find the following. (Include the sign of the charge in your answer.) (a) the charge on the insulating sphere (b) the net charge on the below conducting sphere (c) the charge on the inner surface of the hollow conducting sphere (d) the charge on the outer surface of the hollow conducting sphere

Explanation / Answer

for electric field at 10 cm from the center:

Gauss law states that total electric flux passing through a surface is equal to charge enclosed.

for a concentric sphere of radius 10 cm , charge enclosed= charge on the insulating sphere

as the electric field is radially inward, it means the charge on the sphere is negative.

let charge magnitude be Q.

then using Gauss' Law:

epsilon*electric field*4*pi*distance^2=charge enclosed

==>8.85*10^(-12)*3.15*10^3*4*pi*0.1^2=Q

==>Q=3.5032 nC

so charge on insulating sphere is -3.5032 nC

part b:

at r=50 cm , total charge enclosed =charge on the insulating sphere + charge on the hollow conducting sphere

as electric field is pointed outward, total charge enclosed is positive.

let magnitude be Q.

then 8.85*10^(-12)*235*4*pi*0.5^2=Q

==>Q=6.5337 nC

then charge on hollow sphere=Q-charge on insulator

=6.5337-(-3.5032) nC

=10.037 nC

part c:

as electric field inside a hollow conductor is 0,

total charge enclosed by the inner surface is 0.

hence charge on inner surface=-(charge on insulator)

=3.5032 nC

part d:

charge on outer surface=total charge-charge on inner surface

=6.5337-3.5032

=3.0305 nC

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