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While doing practice question, I came across a question...

A solid sphere of radius a has a cavity of radius b which has a uniformly charge distributed with density $-\rho$ and the remaining part of the sphere has charge density $+\rho$. Find the electric potential at the centre of the sphere of cavity. The centre of the cavity is at a distance c (c>b) from the centre of the sphere.

My approach: I tried to draw this situation. enter image description here

Note that I have putted +q charge inside the cavity so that the charge density on the surface of cavity could be $-\rho$. However I doubt this step taken by me because $\rho$ is generally used for 'charge per unit volume' and I have used it for charge per unit area.

Now charge on the surface of sphere $q_1 = \rho (4\pi a^2)$

Charge on the surface of cavity $q_2 = -\rho (4\pi b^2)$

Now $V_{sphere} - V_{cavity}$ $$\Rightarrow \dfrac{kq_1}{a} - \dfrac{kq_2}{b}$$

$$\Rightarrow \dfrac{k\rho (4\pi a^2)}{a} - \dfrac{-k\rho (4\pi b^2)}{b}$$

$$\Rightarrow \dfrac{\rho (a+b)}{\epsilon_0}$$

But it is wrong. Why is that? What mistake(s) did I commit?

Correct answer is

$\dfrac{\rho}{3\epsilon_0} (\dfrac{3R^2}{2}-\dfrac{2b^3}{c})$

EDIT: After some insight I got a bit near the answer...

$$q_1 = \rho(4/3 \pi (a^3-b^3))$$ $$q_2 = \rho(4/3 \pi b^3)$$ So..$$V_C = \dfrac{\rho}{4 \pi \epsilon_0} (\dfrac{3*4\pi (a^3-b^3)}{2*3a}-\dfrac{4\pi b^3}{3c}) $$ $$= \dfrac{\rho}{3\epsilon_0}(3a^2/2 - \dfrac{3b^3}{2a} -b^3/c)$$

However it still is wrong which means there still a missing concept.:(

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    $\begingroup$ Start with Your very comment that $\rho$ is charge per unit volume and not charge per unit area. $\endgroup$
    – Anubhav Goel
    Commented Feb 6, 2016 at 10:35
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    $\begingroup$ Note the unit (dimension) of your answer! $\endgroup$
    – YiFei
    Commented Feb 6, 2016 at 10:35
  • $\begingroup$ $q_{1} ≠ \rho ( 4πa^2)$ $\endgroup$
    – Anubhav Goel
    Commented Feb 6, 2016 at 10:37
  • $\begingroup$ @AnubhavGoel But at equilibrium position all charges stay at the surface of the conductor $\endgroup$
    – manshu
    Commented Feb 6, 2016 at 10:38
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    $\begingroup$ It is not metal surface that all charges would stay at surface. Charge is uniformly distributed. $\endgroup$
    – Anubhav Goel
    Commented Feb 6, 2016 at 10:40

1 Answer 1

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Potential at center due to +ve sphere is not correct. What you had found is when cavity is at center. However potential due to $-\rho$ is correct.

First consider no cavity

Potential at center of sphere due to uniformly charged complete sphere $ V = 3kq/2a$

Now, potential due to positive charged sphere $cavity$ at center.

$$ V_{1}= \frac{4\rho π b^3}{3c}$$

Subtract it from potential due to uniformly charged complete sphere

$$V - V_{1}$$

Now potential due to negativeness of cavity

$$ V_{2}=- \frac{4\rho π b^3}{3c}$$

Add it to $$V - V_{1}$$.

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  • $\begingroup$ how does it matter if the cavity is not at centre? And how are we supposed to find it? $\endgroup$
    – manshu
    Commented Feb 6, 2016 at 12:16

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