Potential Difference & Force on Electrostatic Sphere

[Silviu]

Homework Statement

A uniformly charged, solid, non-conducting sphere of radius $a$ and charge density $\rho$ has its center located a distance $d$ from a uniformly charged, non-conducting, infinite sheet (charge density $\sigma$). Determine the potential difference between the center of the sphere and the nearest point on the charged sheet. Determine the net electrostatic force on the charged sphere.

Homework Equations

$E=-\nabla V$
$F=qE$

The Attempt at a Solution

I tried something but I want to make sure it is correct. So the electric field created by the plane is $E_{plane}=\frac{\sigma}{2\epsilon}$ and $V_{plane}=\frac{\sigma r}{2\epsilon}$. Using Gauss law the potential inside and outside the sphere is: $V_{sphere}^{inside}=\frac{2\rho r^2}{3\epsilon}$ and $V_{sphere}^{outside}=\frac{Q}{4 \pi \epsilon r}$. So at the center of the sphere the potential is the one created by the plane, so $V_1 = \frac{\sigma d}{2\epsilon}$ and the potential at the surface of the plane is the one created by the sphere $V_2 = \frac{Q}{4 \pi \epsilon d}$. So the potential difference would be just $V_1 - V_2$?
For the second part, the electric field of the plane is constant, so the force is just $\frac{\sigma}{2\epsilon} Q$? With $Q = \rho 4 \pi r^3/3$ Thank you.

 

[TSny]

You haven't been consistent with where you are choosing V to be zero. For the plane, you are choosing V = 0 at the surface of the plane. For inside the sphere, you are choosing V = 0 at the center of the sphere, and for outside of the sphere, you are choosing V = 0 at infinity.

Also, I'm not sure about the factor of 2 in the numerator of the expression for the potential for inside the sphere.

You might consider getting ΔV by integrating the field between points 1 and 2.

I believe your answer for the second part is correct.