Spherical Cavity

1. May 22, 2006

Swagger

A positive charge Q = 1400.00 C is uniformly distributed over the volume of a sphere of radius R = 10.00 m. Suppose a spherical cavity of radius R/2 is cut out of the solid sphere, the center of the cavity being a distance R/2 from the center of the original solid sphere (see figure). The cut-out material and its charge are discarded. What is the magnitude of the electric field produced by this new charge distribution at point P, a distance r = 24.50 m from the center of the original sphere?

(picture attached)

I have calculated the E-field of the point using the equation E=kQ/r^2. I first did this using the radius 24.5m. I then used the radius 24.5m-5m=19.5m. I then subtracted these two E-field calculations and get the wrong answer. E(r) - E(r-R/2)

What am I doing wrong?

Attached Files:

• sphere.gif
File size:
2.1 KB
Views:
172
2. May 22, 2006

siddharth

I can't see the attachments yet, but first of all remember that the electric field is a vector not a scalar. You need to take the directions into account. Then, the electric field given by E=kQ/r^2 is with respect to the center of the sphere in each case. So use the proper distances to the point P. Also, the charge enclosed by your two spheres are different.

Last edited: May 22, 2006
3. May 22, 2006

Swagger

I did. The charge in the center in postitive so the direction on the e-field is positive along the x-axis, right?

4. May 22, 2006

siddharth

Did you account for the fact that the charge on each sphere is different?

5. May 22, 2006

Swagger

I assumed they were the same sense it says "A positive charge is uniformly distributed over the volume of the sphere". If they are not, how do I find out the charges?

6. May 23, 2006

siddharth

It says, the postive charge is uniformly distributed over the volume. So, the volume charge density $\rho$ is constant, and can be calculated.

From this, you can find the charge on the smaller sphere by multiplying by its volume.