NOTE : This problem refers to a diagram on the outline for lecture 2, page 1. Al

Question

NOTE: This problem refers to a diagram on the outline for lecture 2, page 1.

All parts of this problem can be solved using Gauss' law.

Suppose there are two hollow,conducting, concentricspheres, with air between the spheres:
- Sphere 1 (the inner sphere) has inner radiusr1i=7.95cm, outer radiusr1o=9.86cm, and carries chargeq1=-8.35?C.
- Sphere 2 (the outer sphere) has inner radiusr2i=54.2cm, outer radiusr2o=60.2cm and carries chargeq2=3.33?C.
At the common center of the spheres is a charge Q =-7.55?C.

a) Find the charge on each surface.
qon the inner surface of sphere 1:
?C
qon the outer surface of sphere 1:?C
qon the inner surface of sphere 2:?C
qon the outer surface of sphere 2:?C

b) FindE, the magnitude of the electric field, atd=4.0cm from the common center (inside the inner sphere).
E=N/C

c) FindE, the magnitude of the electric field, atd=32.0cm from the common center (between the spheres).
E=N/C

d) FindE, the magnitude of the electric field, atd=114.4cm from the common center (outside the spheres).
E=N/C

e) Suppose the outer sphere is now connected to ground. What will happen to the charge on the inner and outer surfaces of the spheres?

Explanation / Answer

a)

qon the inner surface of sphere 1: +7.55 microC

qon the outer surface of sphere 1: -8.35-7.55 = -15.9 microC

qon the inner surface of sphere 2: +15.9 microC

qon the outer surface of sphere 2: 3.33-15.9 = -12.57 microC

B)

E = kQ/R^2 = -9e9*7.55e-6/(0.04*0.04) = -4.25 * 10^7 N/C

C)

E = k(Q+q1)/R^2 = 9e9*(-7.55e-6-8.35e-6)/(0.32*0.32) = -1.40 * 10^6 N/C

d)

E = k(Q+q1+q2)/R^2 = 9e9*(-7.55e-6-8.35e-6+3.33e-6)/(1.144*1.144) = -8.64 * 10^4 N/C

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