# Spherical Charge distribution

• henrybrent
In summary: If that's all there is to it then that's fine, it doesn't say I have to include it in terms of Q so i'll leave it as P_0, cheers for the... help?

## Homework Statement

A Non-Uniform but spherically symmetric charge distribution has a charge density:

$\rho(r)=\rho_0(1-\frac{r}{R})$ for $r\le R$
$\rho(r)=0$ for $r > R$

where $\rho = \frac{3Q}{\pi R^3}$ is a positive constant

Show that the total charge contained in this charge distribution is Q

## Homework Equations

$Q_{total} = \int \rho(r)dV$ with limits 0 and R
$dV = 4 \pi r^2 dr$

## The Attempt at a Solution

[/B]
I have tried so many solutions it is driving me insane.

Is my dv wrong?

my main method is substituting $\rho_0$ in and then trying to take the constants out of the integral but then I'm stuck with r^3/R or something like that...

This is a 4 mark question, so that usually indicates it's a 4 step process, but this is taking me many steps to get even close..

Last edited:
Would you mind showing more details of your working? I'm afraid I can't really tell what problem you are facing.

$\displaystyle\int^R_0 \rho_0(1- \frac{r}{R}) 4 \pi r^2 dr = \displaystyle\int^R_0\ \frac{3Q}{\pi R^3}(1- \frac{r}{R}) 4 \pi r^2 dr$

That looks reasonable, just carry on evaluating the integral.

Got it!

Q_total = Q that took like 3 hours when it it was, was a very basic integration mistake.

Fightfish said:
That looks reasonable, just carry on evaluating the integral.

I am just wondering, what is the proper definition of r and R?

R is the radius of the sphere? then what is r ?

Fightfish said:
That looks reasonable, just carry on evaluating the integral.

Also, how would I go about deriving an expression for electric field in the region r≤R

henrybrent said:
R is the radius of the sphere? then what is r ?
r is the variable that describes the distance of the point from the origin of the coordinate system. R is just a constant.

henrybrent said:
Also, how would I go about deriving an expression for electric field in the region r≤R
Have you learned about Gauss's law yet?

Fightfish said:
r is the variable that describes the distance of the point from the origin of the coordinate system. R is just a constant.Have you learned about Gauss's law yet?

For a uniform sphere yes I could find the electric field, but not for a non uniform sphere, that has not been taught to us. I have used Gauss' Law = Q_enc/E_0,
I think the integral limits change from 0 to R, to now 0 to r. But I'm not sure

Yup that is correct; because the quantity you want is the charge enclosed within the spherical Gaussian surface that you are using.

Fightfish said:
Yup that is correct; because the quantity you want is the charge enclosed within the spherical Gaussian surface that you are using.
would you be able to guide me a bit further please?

Atm, I have integral of p0(1-r/R)4pi*r^2 dr all divided by e_0 = E4pi*r^2 limits 0 to r

Sorry it's not in latex I have had to type this in a rush!

Well it actually doesn't look like you need a lot of guidance, really. You have all the correct ideas - you just need to have a bit more confidence, sit down and evaluate your integrals!

Fightfish said:
Yup that is correct; because the quantity you want is the charge enclosed within the spherical Gaussian surface that you are using.
Fightfish said:
Well it actually doesn't look like you need a lot of guidance, really. You have all the correct ideas - you just need to have a bit more confidence, sit down and evaluate your integrals!

I still don't get it! If my limits are 0 and r, I am just subbing r into r again for the same thing?I get $\frac{4 \pi*p_0( \frac{r^3}{3} - \frac{r^4}{4R}){\episilom} =E4\pi*r^2 [\itex] Last edited: The r in the integral is a dummy variable, not to be confused with the limit r that defines the size of the Gaussian surface that you have chosen. If you want to be careful, then maybe you want to write $$Q_{enc} = \int_{0}^{r} \rho_{0}\left(1 - \frac{r'}{R}\right) 4 \pi r'^{2} dr'$$ to 'distinguish' the two of them. Fightfish said: The r in the integral is a dummy variable, not to be confused with the limit r that defines the size of the Gaussian surface that you have chosen. If you want to be careful, then maybe you want to write $$Q_{enc} = \int_{0}^{r} \rho_{0}\left(1 - \frac{r'}{R}\right) 4 \pi r'^{2} dr'$$ to 'distinguish' the two of them. [itex] \frac{4 \pi*p_0( \frac{r^3}{3} - \frac{r^4}{4R})}{\epsilon_0} =E4\pi*r^2$

That's where I'm at? not sure how to progress.

Yeah, you're there already, just make E the subject!

Fair enough, I had already done that and then substituted P_0 in and then things got silly. If that's all there is to it then that's fine, it doesn't say I have to include it in terms of Q so i'll leave it as P_0, cheers for the help.

henrybrent said:
Fair enough, I had already done that and then substituted P_0 in and then things got silly.
Silly? How so? Quick way to check your answer is to set r = R, you should recover the expected $\frac{Q}{4\pi R^{2}}$

Fightfish said:
Silly? How so? Quick way to check your answer is to set r = R, you should recover the expected $\frac{Q}{4\pi R^{2}}$

I'm still all over the place. Right,

$E4 \pi r^2 = \frac{\displaystyle \int \rho_0(1-\frac{r}{R})4\pi r^2 dr}{\epsilon_0}$

Can't I just take p_0 out of the integral and then divide by $4\pi r^2$ ?
I'm still not getting the right answer!

The answer you posted earlier, $$\mathbf{E} = \frac{\rho_0( \frac{r}{3} - \frac{r^2}{4R})}{\epsilon_0} \hat{r}$$
(I rearranged and added in the direction vector for completeness)
was correct. So I don't kind of get why you are returning to the integral.

Fightfish said:
The answer you posted earlier, $$\mathbf{E} = \frac{\rho_0( \frac{r}{3} - \frac{r^2}{4R})}{\epsilon_0} \hat{r}$$
(I rearranged and added in the direction vector for completeness)
was correct. So I don't kind of get why you are returning to the integral.

Well I decided to re-write my method more neatly, then it occurred to me, hold on, why can't I just divide by $4\pi r^2$ it saves multiplying out.

Ah, because as I mentioned in an earlier post, the r in the LHS (representing the radius of the Gaussian surface) is different from the r under the integral, which is just a dummy variable. If you want to avoid confusion, then you can do what I suggested by replacing the dummy variable r you are integrating over with r' instead, so as to distinguish the two more clearly.

Fightfish said:
Ah, because as I mentioned in an earlier post, the r in the LHS (representing the radius of the Gaussian surface) is different from the r under the integral, which is just a dummy variable. If you want to avoid confusion, then you can do what I suggested by replacing the dummy variable r you are integrating over with r' instead, so as to distinguish the two more clearly.

Ofcourse! Apologies for the trivial questions! I have not been thinking straight the last few days! thank you for your help.

## What is spherical charge distribution?

Spherical charge distribution refers to the distribution of electric charge on a spherical object or surface. It is a type of electric field that exists in a symmetrical spherical shape around a charged object.

## How is spherical charge distribution calculated?

The spherical charge distribution can be calculated using Gauss's Law, which states that the electric flux through a closed surface is equal to the total enclosed charge divided by the permittivity of free space. This can be represented by the equation E * 4πr² = Q/ε₀, where E is the electric field, r is the radius of the sphere, Q is the total charge, and ε₀ is the permittivity of free space.

## What is the difference between uniform and non-uniform spherical charge distribution?

A uniform spherical charge distribution is when the electric charge is evenly distributed throughout the entire surface of the sphere. This results in a constant electric field at all points on the surface. In contrast, a non-uniform spherical charge distribution is when the electric charge is unevenly distributed, resulting in varying electric fields at different points on the surface.

## What are the applications of spherical charge distribution?

Spherical charge distribution has various practical applications, such as in capacitor design, where spherical capacitors are used to store electric charge. It also plays a role in the study of celestial bodies, as the Earth and other planets can be approximated as spheres with varying charge distributions. Additionally, understanding spherical charge distribution is essential in the field of electrostatics and the design of electrical devices.

## How does spherical charge distribution affect electric potential?

The distribution of electric charge on a sphere affects the electric potential at any point on its surface. The electric potential is directly proportional to the magnitude of the charge and inversely proportional to the distance from the center of the sphere. This means that a point closer to the center of the sphere will have a lower electric potential than a point farther away, and a point with a higher charge density will have a higher electric potential than a point with a lower charge density.