# Ratio of potentials

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1. May 11, 2017

### Buffu

1. The problem statement, all variables and given/known data
Consider a charge distribution which has the constant density $\rho$ everywhere inside a cube of edge $b$ and is zero everywhere outside. Let $\phi$ zero at infinity, $\phi_0$ at the centre of cube and $\phi_1$ at one corner of the sphere. Determine the ratio $\phi_0/\phi_1$. Hint : Think about the potential at the center of the cube with same charge density and with twice the edge length.

2. Relevant equations

3. The attempt at a solution

I don't understand the hint. I guess the potetial at centre would remain if the edge is doubled.

I guess I now have to use superpositon principle but I am not sure how. Please help me understand the hint. :).

2. May 11, 2017

### ehild

From 8 cubes, you can make a cube with twice the edge length. At the center of the big cube, you have eight corners of the small cubes. The potentials add up.

3. May 11, 2017

### Buffu

Oh I am an idiot.

Ok, so if I take some "small cube" of which, when taken 8, makes the given cube. The the 8 corners of the "small cubes" will make the given cube.

So, if potential at corner of each small cube is $\phi$ then $\phi_0 = 8\phi$ and $\phi_0/\phi_1 = 8$ ? is this correct ?

4. May 11, 2017

### haruspex

If the potential at one corner of a small cube, in isolation, is φ then $\phi_0 = 8\phi$, but how does that φ compare with φ1 ?

5. May 11, 2017

### ehild

No, Φ0 and Φ1 refer of the big cube. Φ0=8φ1, where φ1 is the potential at the corner of the small cube. How does the potential at the corners depend on the size of the cube?
How would you calculate the potential at a corner of a cube?

6. May 12, 2017

### Buffu

Any corner of big cube is made up of one corner of small cube, so $\phi_1 = \varphi$ where $\varphi$ is potential of a corner of small cube and $\phi_1$ is potential of corner of large cube. No ?

7. May 12, 2017

### haruspex

No. A cube eight times the volume but the same charge density will have eight times the charge. As against that, the corner will be further from the individual charges on average. Can you figure out how doubling the side of the cube affects the potential at each corner?

8. May 12, 2017

### Buffu

Do I need to do a tedious integration ? I don't think there is any way of figuring that out.

9. May 12, 2017

### haruspex

No need for integration as such.
Consider a tiny cubic element of side dx at distance r from a corner of the whole cube. This generates some potential at the corner. Now double the linear dimensions, so the tiny cube is side 2dx but 2r from the corner, same charge density. What does that do to the potential it generates at the corner?

10. May 12, 2017

### Buffu

Ok. Lets say it generates $\phi$ at the corner. Now we double the side to 2 dx which means we are adding 8 other cubes of side dx all of which contributes $\phi$ to the corner, so the total would be $8\phi$ ?

11. May 12, 2017

### haruspex

No, I don't think you understood my post.
Consider a cube A of side L, and a corner of it, P. Somewhere inside the cube is a small cubic element C of side dx << L. C is distance r from P. The element C has charge ρ dx3. What potential does it raise at P?
Now map this to cube A' of side 2L, corner P', and a small cubic element C', just by doubling all the distances and keeping ρ the same. What is the ratio between the charge on C over that on C'? What is the ratio of the distance from C to P over that from C' to P'? So what is the ratio between the potential C creates at P over that created by C' at P'?

12. May 12, 2017

### Buffu

$\dfrac{\rho dx^3} {r}$ maybe ??

Ratio of charge $\displaystyle {\rho dx^3 \over \rho 8dx^3} = {1\over 8}$

Ratio of distance is $r/2r$ = $1/2$

Ratio of potetial is $\dfrac{\rho dx^3}{r} \over \dfrac{\rho 8dx^3}{ 2r}$ = $2\over 8$ = $1\over 4$.

Sorry for troubling you.

13. May 12, 2017

### haruspex

Right. So what does that give you as the answer to the question?

14. May 12, 2017

### Buffu

Potential at centre is $\phi_0 = 8\varphi$ where $\varphi$ is potential at the corner of small cube.

Also $\displaystyle {\varphi \over\phi_1} = 1/4$, where $\phi_1$ is for corner of large cube.

So $\varphi = \phi/4$ or $\phi_0 = 8 * \phi_1/4 = 2\phi_1$ which is $\phi_0/\phi_1 = 2$ Hope this is correct.

15. May 12, 2017

### haruspex

2 is the answer I got.

16. May 12, 2017

### Buffu

Done Thanks !!