Electric potential of a solid conducting sphere with concentric spherical shell

[kiwikahuna]

Homework Statement

Consider a solid conducting sphere with a radius 1.5 cm and charge -4.4pC on it. There is a conducting spherical shell concentric to the sphere. The shell has an inner radius 3.7 cm and outer radius 5.1 cm and a net charge 27.4 pC on the shell.

A) denote the charge on the inner surface of the shell by Q'2 and that on the outer surface of the shell by Q ''2 . Find the charge Q''2. Answer in units of pC.

B) Find the magnitude of the electric field at point P, midway between the outer surface of the solid conducting sphere and the inner surface of the conducting spherical shell. Answer in units N/C.

C) Find the potential V at point P. Assume the potential at r = infinity. Answer in units of volt.

Homework Equations

E =kQ/r^2
V = kQ/r

The Attempt at a Solution

I've figured out parts A and B but I'm struggling with Part C. I used the equation V = kQ/r where Q = -4.4e-12C ; k = 8.98755e9 and r = 0.026 m
My answer (-1.52 V) is wrong but I have no idea why. Please help if you can.

 

[chanvincent]

We know the total potential is merely the sum of its part, i.e.

$$V_{total}=V_1+V_2+V_3+...$$

In your solution, I can see you have only considered the influence of the center solid conducting sphere, but you have not included the influence made by the spherical shell. In other words, you are missing a term in your solution...

 

[kiwikahuna]

So we are only looking at the charge from the center of the solid conducting sphere and the inner charge of the shell? The inner charge of the shell would be the same charge as the charge from the center except its sign would be opposite. The charge from the center is -4.4e-12 C and the inner charge of the shell is +4.4e-12 C

So should it be...

V1 = (9e9) (-4.4e-12) / 0.026 m
V2 = (9e9) (4.4e-12) / 0.026 m

But then if you add V1 and V2 together, the total potential would be zero?

 

[ice109]

what is the answer? btw kiwikahuna you have a private message

 

[kiwikahuna]

Unfortunately I don't know what the right answer is. Thanks for letting me in on the PM, ice. I hardly ever check it. ^_^