Electric field inside a solid sphere

In summary, the conversation discusses finding the electric field inside a uniformly charged solid sphere using Gauss's law for electric fields. The correct answer is E=\frac{1}{4 \pi \epsilon_0} \frac{q}{R^3} r \hat{r}, which is obtained by considering a concentric Gaussian sphere with radius smaller than R. The interpretation of q in Gauss's law is the amount of charge inside the Gaussian surface. The conversation also mentions using q as the total amount of charge spread across a volume of 4/3 π R3 and finding the charge per unit volume to calculate the total charge within the Gaussian surface. This approach leads to the correct result.
  • #1
roam
1,271
12

Homework Statement



We have a uniformly charged solid sphere whose radius is R and whose total charge is q. I'm trying to find the electric field inside a (r<R).

The correct answer must be:

[itex]E=\frac{1}{4 \pi \epsilon_0} \frac{q}{R^3} r \hat{r}[/itex]

How did they get that answer?

The Attempt at a Solution



Since the sphere is not a shell, the E is not 0. And it should be found by considering a concentric Gaussian sphere with radius smaller than R. So by using Gauss's law for electric fields in the integral form we obtain:

[itex]\oint E . da = E \oint da = E (4 \pi r^2) = \frac{q}{\epsilon_0}[/itex]

[itex]\therefore E = \frac{q}{4 \pi \epsilon_0 r^2}[/itex]

For E outside the sphere we use r>R, and inside the sphere we use r<R. But why is my answer so different from the correct result?

Any help is greatly appreciated.
 
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  • #2
roam said:
[itex]\oint E . da = \frac{q}{\epsilon_0}[/itex]

In Gauss' law as stated above, what is the interpretation of q?
 
  • #3
In the first answer, q is the charge spread out throughout the solid sphere, while in your attempt, q must be the charge enclosed by the gaussian surface.
 
  • #4
TSny said:
In Gauss' law as stated above, what is the interpretation of q?

Thank you very much for the hint! The q in Gauss's law is the amount of charge inside the Gaussian surface. The q in the question is the total amount of charge spread across a volume of 4/3 π R3.

I found the charge per unit volume by q dividing by the volume to get:

[itex]q_{per \ unit \ volume}=\frac{q}{\frac{4}{3} \pi R^3}[/itex]

And to find the total charge

Then I multiplied this by 4/3πr2 to find the total charge in the Gaussian surface of that volume. I got

[itex]q=\frac{qr^3}{R^3}[/itex]

Substituting this back into Gauss's equation I got the right answer. I hope this was the correct thinking and approach.
 
Last edited:
  • #5
Good work :smile:
 
  • #6
Thanks a lot for your help! :)
 

What is an electric field?

An electric field is a physical quantity that describes the force experienced by a charged particle in the presence of other charges. It is represented by a vector and is measured in units of newtons per coulomb (N/C).

How is the electric field inside a solid sphere calculated?

The electric field inside a solid sphere can be calculated using Coulomb's Law, which states that the electric field at a point is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance from the charge.

Does the electric field inside a solid sphere depend on the charge distribution?

No, the electric field inside a solid sphere is independent of the charge distribution. This is because the electric field inside a solid sphere is spherically symmetric, meaning that it has the same magnitude and direction at every point within the sphere.

Can the electric field inside a solid sphere ever be zero?

Yes, the electric field inside a solid sphere can be zero at the center of the sphere. This is because the electric field is proportional to the distance from the center, so at the center (where the distance is zero), the electric field is also zero.

How does the electric field inside a solid sphere change as the radius of the sphere increases?

The electric field inside a solid sphere decreases as the radius of the sphere increases. This is because the electric field is inversely proportional to the square of the distance from the charge, so as the distance increases, the electric field decreases.

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