# Electric field within a sphere

## Homework Statement

A sphere with radius r has uniform charge density ρ within its volume, except for a small hollow sphere located at the center with radius R. Find the electrical field.

ρ=Q/V
∫∫EdS=Q/ε

## The Attempt at a Solution

With the spherical Gaussian surface with radius r:
$E4 \pi r^2=\dfrac{\rho \frac{4\pi r^3}{3}}{\epsilon}$

Is it just a case of finding the electric field from setting a Gaussian surface to the hollow sphere as well and subtracting?

## Answers and Replies

rude man
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You don't need to do any fancy subtractions.

Set your Gaussian surfaces in the hollow part, in the charged part, and outside the charged part. E will differ in all three and will in all cases be a function of the distance form the center.

Thank you for answering, so for a Gaussian surface with radius < R the electric field will be zero since there are no charges in the hollow sphere?

rude man
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Thank you for answering, so for a Gaussian surface with radius < R the electric field will be zero since there are no charges in the hollow sphere?

Yes indeed.

Now, what would be the E field between radii R and r?

E=ρ(Vr-VR)/3εr2?

rude man
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Gold Member
No. Don't involve V. We have area and we have volumetric charge density.

What's the general formula relating the E field, the area of your Gaussian surface and the charge inside that surface?

Oh, and you'll need a new variable for radius since the E field will from now on be a function of the radial distance from the center. Your problem should not have assigned r as the radius of the sphere but too late I guess, so use d as your new variable.

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I've used this since the total charge enclosed in the sphere with radius r is ρVr, which would give me the charge enclosed if there was no hollow sphere, hence why I am subtracting to account for the 'missing charge' due to the hollow sphere.

So E4πd2=Q/ε

I'm sorry for sounding stupid but Gauss' Law is something that just isn't clicking for me.

rude man
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Gauss' law is beautifully simple. It just says that the the total electric flux thru a surface bounding a volume is equal to the charge inside that volume, divided by epsilon.

Flux is the vector dot-product of the electric field times each differential area, integrated over the entire surface. In your case that integral reduces to the area of the surface times the electric field which happens to be constant along the entire surface, due to symmetry.

So now how about we set up a Gaussian surface at a distance d from the center, with R < d < r?

Oh BTW your last formula is correct. What is Q(d)?

So for R<d<r

$d=\sqrt{\dfrac{Q}{4 \pi \epsilon E}}$

and for d>r

$r=\sqrt{\dfrac{Q}{4 \pi \epsilon E}}$

rude man
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So for R<d<r

$d=\sqrt{\dfrac{Q}{4 \pi \epsilon E}}$

Yes, but solve for E. You had the right idea before: Q = charge inside the Gaussian surface at radius = d minus charge of hollow center.
and for d>r

$r=\sqrt{\dfrac{Q}{4 \pi \epsilon E}}$

No. Why change from d to r? But keep in mind that the total charge Q does not increase for d > r.

So just to check that we're agreeing on the labeling.

R is the radius of the hollow sphere
d is the radius of our Gaussian surface
r radius of the charged sphere

$E_{d<r}=\dfrac{Q_d}{4 \pi d^2 \epsilon}$

$E_{d>r}=\dfrac{Q_r}{4 \pi d^2 \epsilon}$

I really appreciate the help.

rude man
Homework Helper
Gold Member
So just to check that we're agreeing on the labeling.

R is the radius of the hollow sphere
d is the radius of our Gaussian surface
r radius of the charged sphere

$E_{d<r}=\dfrac{Q_d}{4 \pi d^2 \epsilon}$

$E_{d>r}=\dfrac{Q_r}{4 \pi d^2 \epsilon}$

I really appreciate the help.

Those are right.

But keep in mind that Qd is the charge inside d, so that formula really applies for all d, by definition.

Qr in your formula is the charge inside r but remember to subtract 4πR2ρ from 4πr2ρ .

Qd in your formula is charge inside d < r but again remember to subtract 4πR2ρ from 4πd2ρ .

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1 person
I'm really grateful for your help, thank you.

Have a nice day/night.