# Isolated charged conducting sphere problem

1. Mar 24, 2005

### abot

an isolated charged conducting sphere with radius 12cm creates an electrical fiels of 4.90^4 21cm away from the center. find the capacitance and charge density?

I used the formula
C= (a*b)/(Ke*(b-a))
but i get the wrong answer what am i doing wrong?

2. Mar 24, 2005

### whozum

If the E Field is 4.9x10^4 21cm away, simplify the charged sphere to a point charge, and calculate Q if you know E using Gauss' Law.

$$\int{E}{dA} = \frac{q}{\epsilon_0}$$ which simplifies to $$EA = \frac{q}{\epsilon_0}$$

Charge Density = Q / A
Whats the surface area of the sphere?

I only know how to do parallel plate capacitances.

3. Mar 24, 2005

### Andrew Mason

The field at 21 cm radius gives you the charge:

$$\int E\cdot dA = 4\pi r^2E = \frac{q}{\epsilon_0}$$

$$q = 4\pi r^2E\epsilon_0$$

Charge density on surface is just q/A.

It may seem surprising but a charged sphere has capacitance. You can pump more charge onto the sphere and build up the voltage on the sphere (relative to ground). This is essentially how the Vandegraff generator works.

Capacitance is defined as:

$$C = q/V = q/\int_R^\infty E dr$$

From Coulomb's law:
$$E = \frac{q}{4\pi\epsilon_0r^2}$$

So from that you can work out the expression for C.

AM

4. Mar 24, 2005

### abot

Andrew Mason, above you say what capacitance is defined as and also tell me coulombs law. after i solve for c, capacitance i get
C= (ab)/(k(a-b))
where a and b are the radius
then when i plug the numbers in, i dont get the right answer. the book gets 13.3pF.
what am i doing wrong?
THANKS

5. Mar 24, 2005

### whozum

$$C = \frac{q}{V} = \frac{q}{\int_R^\infty E dr} = \frac{q}{\int_R^\infty{\frac{q}{4\pi\epsilon_0r^2}}{dr}} = \frac{q}{\frac{-q}{4\pi\epsilon_0R}} = -4\pi\epsilon_0R$$

You know all the variables. Now find the capacitance.

Last edited: Mar 24, 2005