Calculating Electric Field of a Solid Sphere with Uniform Charge Density

  • Thread starter paul11273
  • Start date
  • Tags
    Field
In summary, the conversation is about a HW question involving a solid plastic sphere with uniform charge density and finding the magnitude of the electric field at a given distance. The conversation includes a step-by-step solution to the problem and a suggested alternative method using Gauss's law. The person asking for help is unsure if their solution is correct and is open to feedback and advice. They also mention the possibility of a shorter way to solve the problem.
  • #1
paul11273
156
0
Here is a HW question that I have been wrestling with. Hopefully I have gone about solving it correctly. There is no answer in the back of the text to check my result against. Can anyone have a look and confirm?
Question-
"A solid plastic sphere with radius of 10.0cm has a charge with uniform density throughout it's volume. The electric field 5cm from the center is E=86.0 kN/C radially inward. Find the magnitude of the electric field 15cm from the center."

First I figured I needed to find the charge density. I set EA=qr/e0 ,where e0 is permittivity of free space and qr is total charge at radius of .05m given in the problem. I get to E=pr/3e0 where I am using p as charge density.
I solved for p with the given E and r, so p=4.6*10^-5 (C/m^3).

Next, I found the total charge within the .05m sphere with Q=pV. I got 1.91*10^-7 (C).

Lastly, I used E=ke(Q/r^2) to find the electrical field at radius=.15m. My final result is E=7.63*10^4 (N/C).

Can anyone tell me if I am correct? If not, where did I go wrong? Any help would be appreciated. I feel like I may have done too much work, there is probably a shorter way to the answer.
Thanks.
 
Physics news on Phys.org
  • #2
Originally posted by paul11273
First I figured I needed to find the charge density. I set EA=qr/e0 ,where e0 is permittivity of free space and qr is total charge at radius of .05m given in the problem. I get to E=pr/3e0 where I am using p as charge density.
I solved for p with the given E and r, so p=4.6*10^-5 (C/m^3).
I didn't check your arithmetic, but your equation is correct.
Next, I found the total charge within the .05m sphere with Q=pV. I got 1.91*10^-7 (C).
You need to find the total charge, which extends to radius = 0.1m.
Lastly, I used E=ke(Q/r^2) to find the electrical field at radius=.15m. My final result is E=7.63*10^4 (N/C).
If you had the correct "Q", that method would work.
I feel like I may have done too much work, there is probably a shorter way to the answer.
Here's how I would do this problem. First, I would express all distances in terms of the first radius given, R = 0.05. Thus the radius of the sphere of charge would be 2R; the distance at which you are to find the field would be 3R.

Then I'd apply Gauss's law find the charge density (like you did), but I would express it in terms of the field given (call it E1 and R.

Then I'd find the total charge (like you intended to do) for the sphere of radius 2R. Don't plug in numbers, just get an expression.

Then I'd use Gauss's law again to find the field at a distance of 3R, using the Q I just found.

Do it this way and you'll find that things cancel and you get a simple expression of the field at 3R in terms of the original field at R. The moral: don't be in a rush to plug in numbers and do arithmetic.
 
  • #3
Thanks, Doc Al, for taking the time to review my work.


You need to find the total charge, which extends to radius = 0.1m.

After checking my work I realize that I actually did find the charge of the .1m radius, not the .05m. I must have made an error typing it in. Good catch though.

I will try reworking this problem with your method of using R, 2R and 3R. I understand the concept, but I won't really get it until I work it all out on paper.

Thanks again.
 

FAQ: Calculating Electric Field of a Solid Sphere with Uniform Charge Density

What is Gauss's law and how is it related to electric fields?

Gauss's law is a fundamental law in electromagnetism that relates the electric field to the distribution of electric charges. It states that the electric flux through a closed surface is equal to the total charge enclosed by that surface divided by the permittivity of free space.

How is Gauss's law applied to determine the electric field in a given situation?

To apply Gauss's law, we must first choose a closed surface that encloses the charge distribution of interest. Then, we calculate the electric flux through that surface and equate it to the total charge enclosed divided by the permittivity of free space. This allows us to solve for the electric field at any point on the surface.

Can Gauss's law be used for any charge distribution?

No, Gauss's law can only be used for charge distributions that have spherical, cylindrical, or planar symmetry. This means that the charge must be distributed evenly in a symmetrical manner around a point, line, or plane, respectively.

How does the electric field change as we move farther away from a point charge?

The electric field decreases as we move farther away from a point charge. This is because the electric field is inversely proportional to the square of the distance from the point charge according to Coulomb's law.

Can Gauss's law be used to calculate the electric field inside a conductor?

Yes, Gauss's law can be used to calculate the electric field inside a conductor. This is because the charge distribution inside a conductor is symmetrical, and therefore, Gauss's law can be applied to determine the electric field at any point inside the conductor.

Back
Top