# Charged conducting spheres

1. Mar 3, 2006

### BananaMan

The question i am stuck on and am unable to find anything in my notes about (although i cant find my notes from tuesday however we seem to have covered the topic on friday and maybe just not finished it off in time) is ....

Two isolated spherical conductors one with radius R and one with radius 3R, each carrying a positive charge Qo. The spheres are brought together and then seperated again, find the charge on each sphere.

from what i can think they would act as one surface and the charge would evenly distribute over the whole surface, then i would work out the surface area of each sphere seperately and take a ration and then spread the charge of 2Qo over each sphere and give my answer in terms of Qo?

then a little more difficult

if a small charged sphere is brought into contact with another identical sphere which has no charge then seperated by 10cm the force between them is 9x10^-3N what was the original charge on the first sphere

it relates to my answer from before, if i am right with what i said above, would i work out the difference between the 2 and work out the respective charges, then add them together to get the charge originally on the first?

any and all help greately appreciated, thanks

2. Mar 3, 2006

### Astronuc

Staff Emeritus
Think of electric potential and charge density.

The sphere of radius 3R has 9 times the surface area of a sphere of radius R. If they have the same charge, what can one say about charge density?

3. Mar 3, 2006

### BananaMan

well the formula is charge density = Q/A so the charge density must be 9 times higher on the small sphere, however when they are touched together would this mean that the charge density would try and become equal on both surfaces?

worked through it on this assumption

σ1=9σ2

after

σ=Q/A
σ=2Qo/10=1/5Qo

sub back new value of sigma

1/5Qo*1= 1/5 Qo on small sphere

and therefore 2-1/5 = 9/5 Qo for the large sphere?

if this works then for the second question σafter = 1/2σbefore on both spheres

then using F=(q1q2)/4*pi*Eo*r^2

9x10^-3 * (4*pi*Eo*0.1^2) = 1/2Q * 1/2 Q where Q is the charge on the first sphere originally

solving for Q

1.00x10^-14 = 1/4 Q^2

Q = 2.00x10^-7

that right? thanks again

Last edited: Mar 3, 2006
4. Mar 4, 2006

### Astronuc

Staff Emeritus
Correct. In the first problem, the charge is partitioned according to area.

Sphere 1 would have charge QA1/(A1+A2) and Sphere 2 would have charge QA2/(A1+A2) where Q = 2Qo

In the second problem, the charge is halved on both spheres of the same size.

5. Mar 4, 2006

### BananaMan

thanks, glad i managed to work it out, hope something like that comes up in my exams