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After studying electric field variation inside a non-conducting ball with cavity, I wished to calculate the potential of some points to compare. For simplicity, I made this diagram

here B is point of intersection of cavity and ball and O is centre of ball

And want to calculate the potential at point O (centre of ball) and B (centre of cavity)

Now as my teacher told me, I have to subtract the potential of a point in or outside the cavity from potential of point with respect to the ball

Here we will be taking cavity as another ball of same volume charge density

And used these formulæ

The electrostatic potential $\phi$ observed at radial distance $r$ away from the centre of a non-conducting, uniformly charged ball of radius $R$ and volume charge density $\rho$ is given by $$\phi(r)= \begin{cases} \frac\rho{\epsilon_0}\left(\frac{R^3}{3r}\right)&,r\geqslant R\\ \frac\rho{\epsilon_0}\left(\frac12R^2-\frac16r^2\right)&,r\leqslant R \end {cases}$$

But on applying on this on B and O, I am getting that the potential at O more than at B, which is contrary to the answer key.

So I want any other method to find potential.

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    $\begingroup$ The picture did not show. And if there is a cavity we need to know some relations there too. $\endgroup$ May 3, 2023 at 15:40
  • $\begingroup$ Can you please check now. If there further any mistake please correct it by own. I am not much familiar with latex here $\endgroup$ May 4, 2023 at 2:25
  • $\begingroup$ Contrary to what? @OpenLearner $\endgroup$
    – nasu
    May 4, 2023 at 2:57
  • $\begingroup$ Does the sphere have uniform charge density? $\endgroup$ May 4, 2023 at 3:10
  • $\begingroup$ Hint: You can consider the cavity as a superposition of a positively charged and negatively charged sphere (so that it is neutral). Then, the potential at the center is due to a uniformly (positively) charged large sphere without cavity and a negatively charged sphere placed in the cavity. $\endgroup$ May 4, 2023 at 3:13

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You have stated things inconsistently and I have thus no way to know precisely where you meant to compute.

First of all, I am going to introduce some nomenclature. A ball has insides, and a sphere is just the skin of a ball. This should be consistently used between mathematics and physics. Hence my edit to your post having these changes.

In my current understanding of your problem, the outer big ball has radius $R$ and the cavity has radius $R/2$. Let the centre of the cavity be $A$, contrary to what you wrote, because in your drawing, $B$ is the unique point on the surface of both the ball and the cavity.

The potentials at those points, found by the method you are given, are $$ \phi(O)=\frac5{12}\frac\rho{\epsilon_0}R^2\\ \phi(A)=\frac4{12}\frac\rho{\epsilon_0}R^2\\ \phi(B)=\frac3{12}\frac\rho{\epsilon_0}R^2 $$ i.e. I think your answer key is wrong. Even in the case of just a complete ball and nothing else, the potential just keeps getting stronger as you move towards the centre of the ball. Is the volume charge density $\rho < 0 ?$

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  • $\begingroup$ Yes I checked answer from other sources (teacher) and yes B has potential less than O but I want to know is this correct.. means if I want to calculate potential at these points then the answer you and me calculated is correct or not. $\endgroup$ May 4, 2023 at 15:49
  • $\begingroup$ It is correct. All methods must get the same values because of the same boundary conditions. $\endgroup$ May 4, 2023 at 23:32

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