Electric potential with regards to an insulating spherical shell

[duke91]

An insulating spherical shell with inner radius 25.0 \rm cm and outer radius 60.0 \rm cm carries a charge of + 150.0 \mu {\rm C} uniformly distributed over its outer surface. Point a is at the center of the shell, point b is on the inner surface and point c is on the outer surface.

What will a voltmeter read if it is connected between c and infinity?

I have no idea how to do this question, at all. I am super stuck, can I please get any pointers or instructions how to do this question?

 

[Redbelly98]

Welcome to Physics Forums.

What equations do you have that have something to do with voltage or electric potential?

 

[duke91]

The formulaI know regarding electric potential is: Va-Vb= integral(E* dl)

 

[Redbelly98]

Okay.

Can you find E, and integrate E from point c to infinity?

 

[duke91]

That is where i am stuck... i pretty much get V=q/4pi*epsilon_0r..and then no other clue...

 

[ideasrule]

That is where i am stuck... i pretty much get V=q/4pi*epsilon_0r..and then no other clue...

What else do you need? The question asks for the potential difference and V=q/4pi*epsilon_0r gives it.

 

[duke91]

Well, for that, since it is from infinity..it isf rom infinity to what? when r is equal to infinity the entire thing goes to 0..but the answer is not zero, I have tried it already, what do I use for the 2nd r?

 

[Redbelly98]

The question asks for the potential between infinity and point c. You have correctly used r=∞, the other r will be for point c.

In other words, what is r at point c?

 

[collinsmark]

There's two things about the problem statement that makes this problem easier than it otherwise could be.
(1) The charge is evenly distributed. The key point here is that there is spherical symmetry.
(2) Point C is on the outer surface of the sphere, as opposed to the inner surface, or somewhere else inside.

Gauss' law states that

$$\oint _{S} E \cdot dA = \frac{Q _{enc}}{\epsilon _0}.$$

Due to the spherical symmetry, and due to the fact that point C is on the outer surface (i.e. all the charge in question can be considered to be within the Gaussian surface), it's a fairly easy integral. Use it to obtain E, outside the charged sphere. (Hint: you won't have to actually perform any calculus math to find E. Algebra is all it takes, as long as you know what the surface area of a sphere is.) The beauty of this is that if you're only concerned with point C, you don't need to calculate the electric field anywhere inside the charged sphere (you would have to do that though if you were using the voltmeter to probe points A and/or B though).

Once you have a function for E, you can integrate it to get your potential V, with respect to $$\infty$$. The radius's of interest are r = C and r = infinity. This will require actual calculus, but fortunately the integral isn't too tough.