Electric potential of concentric spheres

[5te4lthX]

Homework Statement

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Homework Equations

V = integral of E * dr

The Attempt at a Solution

I do not completely understand the solution to part B. I was able to solve it with the use of hints.

My guess to the explanation is: the electric field is 0 in the hollow conducting sphere/shell and thus the potential should at the inner surface and the outer surface should be equal.

However, I do not know why the potential is kq/c and not kq/b.

Also, I do not know the solution to part C (the hint says it involves both variables a and b).

My attempt at part C was kq/a

 

[Infinitum]

Since the electric field for b<r<c is zero, charge is uniformly distributed on the outer part of the sphere. So the potential for points B and C(and anything in between) should be...?

Edit: I thought this would be a better way to solve. Do you know Shell Theorem? What does that imply on the potential inside the sphere?

 

[Saitama]

Since E=-dV/dr, and E inside the sphere is zero, therefore -dV/dr is zero and therefore V=constant as the rate of change of potential is zero. Hence, the potential inside the sphere is equal to the potential at the surface of sphere.

 

[ehild]

Homework Equations

V = integral of E * dr

You missed a minus sign, and do not forget that the integration involves an additional constant.
If E=kq/r² the potential is

$V=-\int{Edr}=-\int{k\frac{q}{r^2}}=k\frac{q}{r}+C$

It is very important to note that the potential is continuous. It does not jump at an interface. You need two find the constant C for each domain.

The potential is zero at infinity. That means C=0 for r≥c, V=kq/r for r≥c,so it is V(c)=kq/c at the outer surface of the hollow sphere.

The electric field is zero inside a conductor so the potential is constant.
The potential is continuous, it is the same at the inner side of the outer surface as outside: V=kq/c, and stays the same in the whole shell, even at radius b: V(b)=kq/c.

The potential in the empty space between r=a and r=b is of the form V(r)=kq/r+C' again, with a different constant as in the domain r>c. And the continuity requires that V(b)=kq/b+C'=kq/c at r=b. Find C', and then calculate V(a).

ehild

 

[asteriatic]

- kq/b = kq(b-a)/ab but I do not know what to do with b-a.


I can provide a response to the content regarding the electric potential of concentric spheres. First, the electric potential of a point in space is defined as the amount of work required to move a unit positive charge from infinity to that point. In the case of concentric spheres, the electric potential at any point between the spheres is the sum of the potentials due to each individual sphere.

In part B of the problem, we are given a hollow conducting sphere with charge q on its surface and a smaller solid conducting sphere with radius a and charge -q inside. The electric field inside a hollow conducting sphere is zero, as the charge is distributed uniformly on the outer surface and there is no net field inside. Therefore, the potential at the inner surface of the hollow sphere is the same as the potential at the inner surface of the smaller solid sphere. This potential can be calculated using the equation V = kq/r, where r is the radius of the sphere. Since the potential at the inner surface of the smaller solid sphere is kq/a and the potential at the outer surface of the hollow sphere is kq/b, the potential difference between these two surfaces is kq/b - kq/a = kq(b-a)/ab.

For part C, we are asked to find the total electric potential at a point outside the smaller solid sphere, at a distance r from the center. The electric potential at this point is the sum of the potentials due to the two spheres, which can be calculated using the equation V = kq/r. Therefore, the total potential at this point is kq/a + kq/r = kq(1/a + 1/r). This expression involves both variables a and r, as the potential depends on the distance from the center of the smaller sphere, as well as the radius of the smaller sphere itself.