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Electricity and Magnetism Level 4

A ball of radius R R carries a positive charge whose volume density depends only on separation r r from the ball's centre as ρ = ρ 0 ( 1 − r R ) \rho=\rho_0(1-\frac{r}{R}) where ρ 0 \rho_0 is a constant. Assuming the permittivities of the wall and the environment to be equal to unity find the magnitude of the electric field strength as a function of the distance r r both inside and outside the ball.

If the electric field strength inside the ball is given by E = ρ 0 r A ϵ 0 ( 1 − B r C R ) \large E=\frac{\rho_0r}{A\epsilon_0}(1-\frac{Br}{CR}) and the electric field strength outside the ball is given by

E = ρ 0 R K D ϵ 0 r K − 1 \large E=\frac{\rho_0R^K}{D\epsilon_0r^{K-1}} where A , B , C , D , K A,B,C,D,K are positive integers with B , C B,C coprime. Enter your answer as A + B + C + D + K A+B+C+D+K


The answer is 25.

Shreyansh Mukhopadhyay by Shreyansh Mukhopadhyay , 18, India

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2 solutions

Steven Chase
Jun 13, 2018

Suppose we want to calculate the field at a distance A A from the origin (where A < R A < R ) using the Coulomb method. Gauss works just as well, but there was a question about Coulomb in the comment section.

d q = ρ d V = ρ 0 ( 1 − r R ) 4 π r 2 d r = 4 π ρ 0 ( r 2 − r 3 R ) d r d q = \rho \, dV = \rho_0 (1 - \frac{r}{R}) \, 4 \pi \, r^2 \, d r = 4 \pi \rho_0 (r^2 - \frac{r^3}{R}) \, d r

The electric field contribution is (since each shell acts as an equivalent point charge at the origin):

d E = 1 4 π ϵ 0 d q A 2 = ρ 0 ϵ 0 A 2 ( r 2 − r 3 R ) d r d E = \frac{1}{4 \pi \epsilon_0} \frac{d q}{A^2} = \frac{\rho_0}{\epsilon_0 A^2} (r^2 - \frac{r^3}{R}) \, d r

The total field at A is (neglect contributions for r > A r > A ):

E = ρ 0 ϵ 0 A 2 ∫ 0 A ( r 2 − r 3 R ) d r = ρ 0 ϵ 0 A 2 ( A 3 3 − A 4 4 R ) = ρ 0 ϵ 0 ( A 3 − A 2 4 R ) = ρ 0 A ϵ 0 ( 1 3 − A 4 R ) = ρ 0 A 3 ϵ 0 ( 1 − 3 A 4 R ) E = \frac{\rho_0}{\epsilon_0 A^2} \int_0^A (r^2 - \frac{r^3}{R}) \, d r = \frac{\rho_0}{\epsilon_0 A^2} \Big ( \frac{A^3}{3} - \frac{A^4}{4 R} \Big ) \\ = \frac{\rho_0}{\epsilon_0 } \Big ( \frac{A}{3} - \frac{A^2}{4 R} \Big ) \\ = \frac{\rho_0 \, A}{\epsilon_0 } \Big ( \frac{1}{3} - \frac{A}{4 R} \Big ) \\ = \frac{\rho_0 \, A}{3 \epsilon_0 } \Big ( 1 - \frac{3 A}{4 R} \Big )

Now consider the case for A > R A > R .

d E = ρ 0 ϵ 0 A 2 ( r 2 − r 3 R ) d r d E = \frac{\rho_0}{\epsilon_0 A^2} (r^2 - \frac{r^3}{R}) \, d r

Integrating:

E = ρ 0 ϵ 0 A 2 ∫ 0 R ( r 2 − r 3 R ) d r = ρ 0 ϵ 0 A 2 ( R 3 3 − R 3 4 ) = ρ 0 ϵ 0 A 2 R 3 12 E = \frac{\rho_0}{\epsilon_0 A^2} \int_0^R (r^2 - \frac{r^3}{R}) \, d r = \frac{\rho_0}{\epsilon_0 A^2} \Big ( \frac{R^3}{3} - \frac{R^3}{4} \Big ) = \frac{\rho_0}{\epsilon_0 A^2} \frac{R^3}{12}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Ah, I took 'A' as variable (r). Thanks a lot sir.

Shreyansh Mukhopadhyay - 2 years, 12 months ago

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Yeah, I knew I would do the same if I kept them the same.

Steven Chase - 2 years, 12 months ago
Md Zuhair
Jun 13, 2018

Bro I have a small but cute doubt. See if you do the first part using Gauss you get different answer and if you do with Kdq/r^2 integral. We get another different answer... Why is it so?

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Actually I too did that mistake and came up with ρ r ϵ 0 ( 1 − r 2 R ) \large\frac{\rho r}{\epsilon_0}(1-\frac{r}{2R}) but when I did it with Gauss' Law got the correct answer. I couldn't spot the mistake. Sir @Steven Chase , please help.

Shreyansh Mukhopadhyay - 2 years, 12 months ago

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