A solid insulating sphere of radius a = 4.8 cm is fixed at the origin of a co-or

Question

A solid insulating sphere of radius a = 4.8 cm is fixed at the origin of a co-ordinate system as shown. The sphere is uniformly charged with a charge density = -302 C/m3. Concentric with the sphere is an uncharged spherical conducting shell of inner radius b = 11.9 cm, and outer radius c = 13.9 cm.

1)

What is Ex(P), the x-component of the electric field at point P, located a distance d = 27 cm from the origin along the x-axis as shown?

N/C

2)

What is V(b), the electric potential at the inner surface of the conducting shell? Define the potential to be zero at infinity.

V

3)

What is V(a), the electric potential at the outer surface of the insulating sphere? Define the potential to be zero at infinity.

V

4)

What is V(c) - V(a), the potentital differnece between the outer surface of the conductor and the outer surface of the insulator?

V

5)

A charge Q = 0.036C is now added to the conducting shell. What is V(a), the electric potential at the outer surface of the insulating sphere, now? Define the potential to be zero at infinity.

Explanation / Answer

A, B, and C are radiuses

1) (rho)(pi)(4/3)(a^3) = Q

Ex = (KQ) / r^2

2) V = (KQ) / C

3) you have to integrate these but i'll give you the final form

KQ / r ] from infinity to C + KQ / r ] from b to a

4) integrate from infinity inwards

(KQ / C) - (Answer from 3)

5) make sure to adjust Q accordingly

K(Qtotal) /r ] from infinity to c + KQ/r] from c to b + KQ/r] from b to a

-remember for for R greater than or equal to c we will use new charge
-at radius b we will use old charge
-at radius a we will use new charge
-you will use both charges for the second integral

- all of these will cancel so you can do K(newQ) / r ] at radius A.

- these are already integrated just simply plug in.

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