# Force on a sphere due to a conducting plate

#### SingDingLing

Homework Statement
A sphere of radius R is situated a distance d away from an infinite, grounded conducting sheet. (d is the distance from the center to the plate, and d>R.) If the sphere has a uniform charge density p, calculate the force on the sphere.
Homework Equations
F=Integral(E dV) over the volume of the sphere
I tried to find the the Electric field due to the image charge. So the potential due to the image charge is V=-(pR^2)/√(4R^2-4rRcos(θ)+r^2). When I took the gradient of that in spherical coordinates, I got a mess that doesn't seem to be possible to integrate.

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#### kuruman

Homework Helper
Gold Member
Can't you use Gauss's Law?

#### SingDingLing

Can't you use Gauss's Law?
I believe that Gauss's law would be another way to find that electric field due to the image charge (well technically the conducting plate), but the problem is still that it seems to come to an integral that doesn't seem doable.

#### kuruman

Homework Helper
Gold Member
If the "real" sphere has radius $R$ and uniform charge distribution $\rho$, the image sphere will also have radius $R$ and uniform charge distribution $-\rho$. Why can you not use Gauss's Law to find the electric field in this spherically symmetric case? In fact you don't even need the electric field or the potential. Just ask yourself the question, what is the force of attraction between two spheres of uniform density $\rho$ and $-\rho$ and radius $R$ with center-to-center separation $2d$? The electric potential in the region of the real sphere is the same as the electric potential if the plane is replaced by the image sphere in which case the force on the real sphere will be the same. That's the beauty of the image charge approach.

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#### SingDingLing

At first, I thought I could just use Coulomb's law to find the force of attraction between the spheres. But then I was told that since the sphere has a uniform charge density, it isn't the same as Coulomb's law between two point charges with the same charge as the spheres. Hence, I tried using the gradient of the potential from the image sphere.

#### haruspex

Homework Helper
Gold Member
2018 Award
I was told that since the sphere has a uniform charge density, it isn't the same as Coulomb's law between two point charges with the same charge as the spheres.
Isn't the spherically symmetric charge case the one situation in which you can treat the sphere as a point charge (for regions outside the sphere)?

"Force on a sphere due to a conducting plate"

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