Electric Potential Homework: Find Potential of Middle Charge

[atomqwerty]

Homework Statement

Three conductor spheres (radii R) are located aligned with their centers separated by the same distance d (d>>R). The speheres at both sides are connected to ground, and the middle one is given a charge Q. Find the potential of this last one.

Homework Equations

E = -Grad V
∫EdS = Q/e0 (Gauss) [e0: permitivity vacuum]

The Attempt at a Solution

I do know the very solution, but I don't know how to reach there. If a calculate E and then the potential V for the middle charge, I don't get the solution. The thing is... conductors at both sides has got potential V=0 (ground state) and so that those don't contribute to the general potential.Thanks

 

[anathema]

for posting this interesting problem! Let's work through it together.

First, let's draw a diagram to make sure we understand the setup correctly. We have three conductor spheres, each with a radius R, aligned with their centers separated by a distance d. The spheres on the left and right are connected to ground, and the middle one is given a charge Q. Is that correct?

Now, let's think about what we know and what we need to find. We know the distance between the spheres (d) and the radius of the spheres (R). We also know the charge on the middle sphere (Q). What we need to find is the potential of the middle sphere.

To find the potential, we can use the formula V = -∫E•dr, where E is the electric field and dr is a small displacement in the direction of the electric field. We can also use the equation ∫EdS = Q/ε0 (Gauss's law), where ε0 is the permittivity of vacuum.

Let's start by finding the electric field at a point on the surface of the middle sphere. Since the two outer spheres are grounded, we know that the potential at their surfaces is 0. This means that the electric field lines will terminate on their surfaces, perpendicular to the surface. Since the spheres are aligned, the electric field at a point on the surface of the middle sphere will be the same as the electric field at a point on the surface of the outer spheres. We can use Gauss's law to find the electric field at a point on the surface of the outer spheres, and then we can use the principle of superposition to find the electric field at a point on the surface of the middle sphere.

Using Gauss's law, we can write ∫EdS = Q/ε0 as ∫E•dS = Q/ε0. Since the electric field is constant and perpendicular to the surface, we can pull it out of the integral. This gives us E∫dS = Q/ε0. The integral of dS over the surface of a sphere is just 4πR^2, so we can write this as E(4πR^2) = Q/ε0. Solving for E, we get E = Q/(4πε0R^2).

Now, let's use the principle of superposition to find the electric field at a point on the surface of the middle sphere