# Calculate the electric potential of a sphere

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1. Nov 6, 2016

### doktorwho

1. The problem statement, all variables and given/known data
A metal sphere of radius $a=1cm$ is charged with $Q_a=1nC$. Around a sphere is placed a spherical shell of inner radius $b=2cm$ and outer radius $c=3cm$. The electrical potential of the shell in refenrence to a point in the infinity is $V=150V$. The spheres are in a vacuum. Calculate:
a) The electric potential of the inner sphere in reference to a point in the infinity
b) The total charge of the spherical shell
2. Relevant equations
3. The attempt at a solution

Im gonna give you the results straight away so you can help me faster.
a) $600V$
b) $-0.5nC$
The first one i dont know how to start but the b) part i tried like this:
$V=\frac{Q}{4πξ_oR}$ where $R=C$ and got 0.5 but i dont get the $-$ part. I dont understand this problem..Can you help?

2. Nov 6, 2016

### I like Serena

Hi doktorwho!

For (b) you calculated the total enclosed charge, which is indeed +0.5 nC. Since the inner sphere carries 1 nC, the outer spherical shell must have -0.5 nC.

3. Nov 6, 2016

### hilbert2

If the sphere and the shell are made of metal (a conductor), the electrical potential is constant in them. Use Gauss's law to determine the electric field at different radii between the sphere and the outer shell, and then integrate to find the potential difference between the sphere and the shell.

4. Nov 7, 2016

### doktorwho

Can i do it like this?:
$V_a+∫Edl+V_c=V_{inner}$
Basically im adding up the potential of the sphere at surface, the potential difference between point b and point a and the potential of the whole. Simply:
$V_b+V_c=V_{inner}$ where i use the respective Q-s. Since the potential of the $V_c=\frac{0.5}{4πε_or_c}$ and the $V_b$ is proportional to $V_c$ $V_b=2*\frac{3}{2}*V_c=450V$ i get 600. Is this correct thinking?

5. Nov 7, 2016

### hilbert2

^ yes, that seems to be correct.

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