Electric field of sphere with spherical cavity cut out

[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?

 

[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.

 

[Swagger]

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

 

[siddharth]

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

 

[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?

 

[siddharth]

It says, the positive 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.