Ratio of Potential: Calculating ##\phi_0/\phi_1##

[Buffu]

Homework Statement

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.

Homework Equations

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. :).

 

[ehild]

Homework Statement

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 cube. 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.

Homework Equations

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. :).

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.

 

[Buffu]

sphere.

Oh I am an idiot.

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.

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 ?

 

[haruspex]

if potential at corner of each small cube is ϕ then $\phi_0 = 8\phi$

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 ?

 

[ehild]

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 ?

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?

 

[Buffu]

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?

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 ?

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 ?

 

[haruspex]

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 ?

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?

 

[Buffu]

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?

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

 

[haruspex]

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

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?

 

[Buffu]

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?

Ok. Let's 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 $8\phi$ ?

 

[haruspex]

Ok. Let's 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 $8\phi$ ?

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'?

 

[Buffu]

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?

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

No, I don't think you understood my post.
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'?

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.

 

[haruspex]

$\dfrac{\rho dx^3} {r}$ 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.

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

 

[Buffu]

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

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.

 

[haruspex]

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.

2 is the answer I got.

 

[Buffu]

2 is the answer I got.

Done Thanks !