# Electric field inside a non uniform spherical distribution.

## Homework Statement

Let ρ(r)= Qr/πR4 be the charge density distribution for a solid sphere of radius R and total charge Q.For a point 'P' inside the sphere at distance r1 from the centre of the sphere, the magnitude of electric field is ?
A) 0
B) Q/4πε°(r1)2
C) Q(r1)2/4πε°R4
D) Q(r1)2/3πε°R4

## The Attempt at a Solution

Constructing a spherical gaussien surface on which point P lies,
Φ= EΔs=q/ε°
E4π(r1)2= q/ε°
where q is the charge enclosed by the gaussien surface.
I'm having trouble as I don't know how to find q. I know dq= ρdv and so q=∫ ρdV , but if I do substitute ρ in that integral , I don't know what to do with the dV.
Please give me a hint as to how to proceed.

Delta2
Homework Helper
Gold Member
Have you been taught what is the expression for the volume element ##dV## in spherical coordinates?

Have you been taught what is the expression for the volume element ##dV## in spherical coordinates?
No, we were not taught anything like that. Is that the only way, whatever it is?

SammyS
Staff Emeritus
Homework Helper
Gold Member
Have you been taught what is the expression for the volume element ##dV## in spherical coordinates?
No, we were not taught anything like that. Is that the only way, whatever it is?
You don't need the general expression for the volume element ##\ dV \ ## in spherical coordinates.

The distribution has spherical symmetry, so you can find ##\ dV \ ## as follows.

A sphere of radius, ##\ r\,,\ ## has a surface area of ##\ 4\pi r^2\ ## so a spherical shell of radius, ##\ r\,,\ ## and thickness, ##\ dr\,,\ ## has a volume of ##\ dV=4\pi r^2dr\ .##

Use that for ##\ dV \ .##

takando12, Delta2 and conscience
Delta2
Homework Helper
Gold Member
Ok, well SammyS suggestion saves you from extra work and keeps you inside what you already know without the need for the general expression for dV.

However , my egoism and jealousy (oO?!?) (the most common traits of all human beings regardless of race, culture, religion or whatever) makes me to say that my suggestion is not wrong, it is correct also, it just requires some extra work...

However , my egoism and jealousy (oO?!?) (the most common traits of all human beings regardless of race, culture, religion or whatever) makes me to say that my suggestion is not wrong, it is correct also, it just requires some extra work...
Your suggestion is correct .But SammyS 's suggestion should invariably be the first line of approach (easier and simpler) .

I am Ok with jealousy part , but the egoism part has forced me to write this reply .

Honestly , I was really surprised when I read your initial suggestion to the OP of using "spherical coordinates" . I was even more surprised to read this
At the moment cant think of other way on how to integrate to get the enclosed charge
.

I have read few of your responses on PF and I think you are pretty good . Your knowledge is far superior to mine . But I have a couple of suggestions .

1) When anybody decides to post in a Homework Help thread , he/she should be confident of providing complete assistance . Because as soon as the first response is made , that takes the thread out of the radar of almost every experienced HH ( by taking it out the unanswered thread list) . The probability of any other HH chipping in becomes very less . So if the OP does not get proper assistance at his required level , the poor guy is left in a lurch . I have seen this here at PF .

2) Even if somebody knows advanced stuff , the first level of attacking the problem should always be something simpler ( if possible) .

I hope the common traits of a human are not interfering in acknowledging the above points .

takando12 and Delta2
Delta2
Homework Helper
Gold Member
yes, point 1) is pretty strong but (egoism interfering) it doesn't mean HHs know better than ME, they just know how to help better :D

takando12 and conscience
yes, point 1) is pretty strong
Thanks for acknowledging this point . Not many do .

but (egoism interfering) it doesn't mean HHs know better than ME, they just know how to help better :D
Absolutely .

You don't need the general expression for the volume element ##\ dV \ ## in spherical coordinates.

The distribution has spherical symmetry, so you can find ##\ dV \ ## as follows.

A sphere of radius, ##\ r\,,\ ## has a surface area of ##\ 4\pi r^2\ ## so a spherical shell of radius, ##\ r\,,\ ## and thickness, ##\ dr\,,\ ## has a volume of ##\ dV=4\pi r^2dr\ .##

Use that for ##\ dV \ .##
Yes, that is within my school level jurisdiction.
It works and I got option c) which is right.

Thank you,
@SammyS and @Delta² for the legal and illegal solutions respectively. Alas, we don't learn lots of beautiful things in school.
and @conscience for highlighting the daily horrors of the OP. We sometimes get petrified when we hear completely fresh concepts that we have never encountered before.
my egoism and jealousy (oO?!?) (the most common traits of all human beings regardless of race, culture, religion or whatever)
Long live the egotist.

Delta2