An insulating sphere of radius 2 cm and charge -5 C is surrounded by a spherical

Question

An insulating sphere of radius 2 cm and charge -5 C is surrounded by a spherical metallic shell of inner radius 4 cm and outer radius 6 cm. The outer shell has a net charge of 3 C . You may assume that charge on the insulator is distributed uniformly throughout its volume.

Part A

What is the electric field for r < 2.0 cm? Give your answer in terms of the variable r.

Part B

What is the electric field for 2 cm < r < 4 cm? Give your answer in terms of the variable r.

Part C

What is the electric field for 4 cm < r < 6 cm? Give your answer in terms of the variable r.

E=.....

Part D

What is the electric field for r > 6 cm?

E=.....

Part E

What is the electric potential at r = 7 cm?

V=.....

Part F

What is the electric potential at r = 5 cm?

V=.....

Part G

What is the electric potential at r = 3 cm?

V=.....

Part H

What is the electric potential at r = 0 cm?

V=.....

Explanation / Answer

Part A

charge in the insulated sphere is q = -5 C , radius of the sphere is R = 2 cm

charge perunit volume is = q/V, where V = (4/3)R3

Let us draw a spherical Gaussian surface of radius r within the sphere. Then charge enclosed within the sphere of radius r will be

q1 = V1, where V1 = (4/3)r3

Let the field at r be E

Using Gauss's law, we have

E.(4r2) = q1/0

or, E = (1/40)q1/r2 = (1/40)V1/r2 = (1/40)(q/V)V1/r2

or, E =  (1/40)qr3/r2R3 = (1/40)qr/R3 = (9X109N-m2/C2)(-5 C)r/(0.02m)3

or, E = -5.625X109r, (negative sign shows that the field is radially inward)

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Part B

Again, draw a spherical Gaussian surface of radius r in this region. Then charge enclosed within the sphere of radius r will be q = -5 C

Let the field at r be E

Using Gauss's law, we have

E.(4r2) = q/0

or, E = (1/40)q/r2 = (9X109N-m2/C2)(-5 C)/r2

or, E = -4.5X104/r2 (negative sign shows that the field is radially inward)

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Part C

The region 4 cm < r < 6 cm lies within the metallic shell. So field will be zero.

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Part D

for r > 6cm

The chagre enclosed in the Gaussian sphere is q = -5 C + 3 C = -2 C

So Using Gauss's law, we have

E.(4r2) = q/0

or, E = (1/40)q/r2 = (9X109N-m2/C2)(-2 C)/r2

or, E = -1.8X104/r2 (negative sign shows that the field is radially inward)

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