Force on a sphere due to a conducting plate

In summary, the electric field due to the image charge is V=-(pR^2)/√(4R^2-4rRcos(θ)+r^2). However, it's difficult to integrate this field and Gauss's law would be another way to find this field.
  • #1
SingDingLing
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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.
Relevant 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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  • #2
Can't you use Gauss's Law?
 
  • #3
kuruman said:
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.
 
  • #4
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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  • #5
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.
 
  • #6
SingDingLing said:
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)?
 

FAQ: Force on a sphere due to a conducting plate

1. What is the force on a sphere due to a conducting plate?

The force on a sphere due to a conducting plate is known as the electrostatic force. This force is caused by the electric charges on the conducting plate, which create an electric field that exerts a force on the charged sphere.

2. How is the force on a sphere calculated?

The force on a sphere due to a conducting plate can be calculated using Coulomb's law, which states that the force is equal to the product of the charges divided by the square of the distance between them. In this case, the charges are the charge on the sphere and the charge on the conducting plate.

3. What factors affect the force on a sphere due to a conducting plate?

The force on a sphere due to a conducting plate is affected by the magnitude of the charges on the sphere and the plate, as well as the distance between them. Additionally, the presence of other charged objects in the vicinity may also influence the force.

4. What is the direction of the force on a sphere due to a conducting plate?

The direction of the force on a sphere due to a conducting plate is towards the conducting plate, as the electric field lines always point towards the plate. This force is attractive if the charges on the sphere and the plate are of opposite signs, and repulsive if they are of the same sign.

5. Can the force on a sphere due to a conducting plate be negative?

Yes, the force on a sphere due to a conducting plate can be negative if the charges on the sphere and the plate are of opposite signs. In this case, the force will be attractive, pulling the sphere towards the plate. However, if the charges are of the same sign, the force will be positive and repulsive, pushing the sphere away from the plate.

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