# What is the charge on the shell? Calculate the potential on the surface of the shell.

I need help with my physics homework…

“Consider a charged metallic spherical shell of a radius a = 1.75 cm. The potential at a distance of r = 6.0 cm from the center of the shell is measured to be 1.5 kV. What is the charge on the shell? Calculate the potential on the surface of the shell.”

I know the formula that must be used is V = kQ/r. Thus, V is given (1500 V), r is given (0.06 m), and k is a constant (8.99×10^9 N.m^2/C^2).

I’m a bit puzzled on how to solve the problem though…could someone help me understand solving for potential?

Observing members:
0
Composing members:
0
## 8 Answers

Hm….

Is “Q” the charge? I believe it is…so if this is the charge, then I simply use algebra to figure out what Q is because everything else is given.

However, now I’m not quite sure what the difference between, “What is the charge on the shell?” and “Calculate the potential on the surface of the shell.”

Unless if it the same question?

@sillymichelleyoung

Charge is measured in Coulombs (Q), and potential is measured in Volts. It’s unclear which quantity the OP is looking for. Perhaps it is both?

@koanhead – I figured that it’s possible to calculate the charge (Q) because all of the other variables are given, but what I’m trying to figure out is the “Calculate the potential on the surface of the shell.”

…Unless…if the given V is for the inside of the shell, then I would have to calculate..

ahh I don’t know >~>

@sillymichelleyoung

The given V (1500V) is the value at 6.0 cm from center, the surface of the shell is at 1.75 cm.

V=(kQ)/r

Vr=kQ

(Vr)/k=Q

Plug in some numbers and voila!

Wait…sooooo to find the charge on the shell as @jerv pointed out would be,

Q = (1500)(0.06)/8.99×10^9

Now, I was thinking about what it meant by “from the surface of the shell” and as @koanhead said it would be 1.75. So, would that mean to find the potential on that, I would simply use V = kQ/R where k is still the constant, Q is the charge from the previous post because the charge would be the same, and R is 0.0175?

Ahhh, I think I finally understand, thanks!

If you understand now, then you have learned, and isn’t that the whole point?

## Answer this question

This question is in the General Section. Responses must be helpful and on-topic.