# Calculating Electric Potential and Energy in a System of Spherical Conductors

• lorenz0
In summary: You also have to account for the fact that the potential energy stored in the electric field between ##C_1## and ##C_2## (before the connection) changes when the connection is made. You are right that you don't have to compute the initial and final energies separately, because this is essentially done automatically by the way you are setting up the problem. But you do have to understand what is going on and what the meaning of the final answer is. Also, think about whether this is a general result, or whether it relies on the special geometry of this particular three-conductor problem.
lorenz0
Homework Statement
Three spherical thin and hollow conductors ##C_1, C_2, C_3,## have radii respectively ##R_1, R_2, R_3.## A charge ##q_1## is put on ##C_1,## a charge ##q_2## is put on ##C_2,## and a charge ##q_3## on ##C_3.##
Find: (a) the electric field ##E## at a point ##P## at distance ##d## from ##R_3;## (b) The potential difference ##V_3 -V_1## between the conductors ##C_3## and ##C_1;## (c) Now, the conductors ##C_1## and ##C_2## are joined by a conducting cable. Find the variation in electrostatic energy energy ##\Delta U_e.##
Relevant Equations
##V_3-V_1=\int_{R_3}^{R_1}\vec{E}\cdot d\vec{l},\ U=\frac{1}{2}\int \sigma V da##
(a) Using Gauss's Law ##E_P=\frac{q_1+q_2+q_3}{4\pi\varepsilon_0(R_1+R_2+R_3+d)^2};(b) V_3-V_1=\int_{R_3}^{R_2}\frac{q_1+q_2}{4\pi\varepsilon_0 r^2}dr+\int_{R_2}^{R_1}\frac{q_1}{4\pi\varepsilon_0 r^2}dr=\frac{q_2}{4\pi\varepsilon_0}\left(\frac{1}{R_3}-\frac{1}{R_2}\right).##

(c) ##U_i=\frac{1}{8\pi\varepsilon_0} \left(\frac{q_1^2}{R_1}+\frac{q_2^2}{R_2}+\frac{q_3^2}{R_3}\right)## and when ##C_1## and ##C_2## are connected we have that ##q_{1f}+q_{2f}=Q##, where ##Q:=q_1+q_2## and ##V_1=V_2\Leftrightarrow \frac{q_1}{4\pi\varepsilon_0 R_1}=\frac{q_2}{4\pi\varepsilon_0 R_2}## which together imply that ##q_{1f}=\frac{R_1}{R_1+R_2}Q,\ q_{2f}=\frac{R_2}{R_1+R_2}Q## so ##U_f-U_i=\frac{1}{4\pi\varepsilon_0 R_1}\left(q_{1f}^2-q_1^2\right)+\frac{1}{4\pi\varepsilon_0 R_2}\left(q_{2f}^2-q_2^2\right)##

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lorenz0 said:
(a) Using Gauss's Law ##E_P=\frac{q_1+q_2+q_3}{4\pi\varepsilon_0(R_1+R_2+R_3+d)^2}##

You are not expressing the distance from the center of the system to the field point correctly.

lorenz0 said:
(b) ##V_3-V_1=\int_{R_3}^{R_2}\frac{q_1+q_2}{4\pi\varepsilon_0 r^2}dr+\int_{R_2}^{R_1}\frac{q_1}{4\pi\varepsilon_0 r^2}dr=\frac{q_2}{4\pi\varepsilon_0}\left(\frac{1}{R_3}-\frac{1}{R_2}\right).##

You have set this up correctly in terms of the integrals, but your result for the evaluation of the integrals is not correct.

lorenz0 said:
(c) ##U_i=\frac{1}{8\pi\varepsilon_0} \left(\frac{q_1^2}{R_1}+\frac{q_2^2}{R_2}+\frac{q_3^2}{R_3}\right)##
This is not the correct expression for the initial energy of the system. Energy does not obey a superposition principle.

##\frac{1}{8\pi\varepsilon_0} \frac{q_1^2}{R_1} ## is the energy associated with conductor ##C_1## if the other conductors are not present. Similarly for the other two conductors. But, when all three conductors are present, the total energy is not the sum of these individual energies.

lorenz0 said:
and when ##C_1## and ##C_2## are connected we have that ##q_{1f}+q_{2f}=Q##, where ##Q:=q_1+q_2## and ##V_1=V_2\Leftrightarrow \frac{q_1}{4\pi\varepsilon_0 R_1}=\frac{q_2}{4\pi\varepsilon_0 R_2}## which together imply that ##q_{1f}=\frac{R_1}{R_1+R_2}Q,\ q_{2f}=\frac{R_2}{R_1+R_2}Q##
When ##C_1## and ##C_2## are connected so that ##V_1=V_2##, can there be any electric field between ##C_1## and ##C_2##? What does this tell you about the final charge on ##C_1##?

lorenz0
TSny said:
You are not expressing the distance from the center of the system to the field point correctly.
You have set this up correctly in terms of the integrals, but your result for the evaluation of the integrals is not correct.This is not the correct expression for the initial energy of the system. Energy does not obey a superposition principle.

##\frac{1}{8\pi\varepsilon_0} \frac{q_1^2}{R_1} ## is the energy associated with conductor ##C_1## if the other conductors are not present. Similarly for the other two conductors. But, when all three conductors are present, the total energy is not the sum of these individual energies.When ##C_1## and ##C_2## are connected so that ##V_1=V_2##, can there be any electric field between ##C_1## and ##C_2##? What does this tell you about the final charge on ##C_1##?
Thanks!
For part (a) it should have been ##E_P=\frac{q_1+q_2+q_3}{4\pi\varepsilon_0 (R_3+d)^2}##.
For part (b) it should have been ##V_3-V_1=\frac{q_1}{4\pi\varepsilon_0}\left(\frac{1}{R_3}-\frac{1}{R_1}\right)+\frac{q_2}{4\pi\varepsilon_0}\left(\frac{1}{R_3}-\frac{1}{R_2}\right).##

For part (c) since ##C_1## and ##C_2## have the same potential the electric field between them has magnitude 0, so I was thinking that maybe the difference in energy relative to the initial situation is the one that was stored in the electric field between them, which should be ##\frac{\varepsilon_0}{2}\int_{R_1}^{R_2}E^2 dV##. This approach would also have the additional benefit of not having to compute the initial and final energy, but does it make sense?

Last edited:
For part (c) I think the problem is asking you to find the electrostatic energy before and after the connection is make and take the difference After - Before. By the way, there is a factor of ##\frac{1}{2}## missing in front of the integral ##\varepsilon_0\int_{R_1}^{R_2}E^2 dV.##

lorenz0
lorenz0 said:
For part (a) it should have been ##E_P=\frac{q_1+q_2+q_3}{4\pi\varepsilon_0 (R_3+d)^2}##.
Yes

lorenz0 said:
For part (b) it should have been ##V_3-V_1=\frac{q_1}{4\pi\varepsilon_0}\left(\frac{1}{R_3}-\frac{1}{R_1}\right)+\frac{q_2}{4\pi\varepsilon_0}\left(\frac{1}{R_3}-\frac{1}{R_2}\right).##
Yes

lorenz0 said:
For part (c) since ##C_1## and ##C_2## have the same potential the electric field between them has magnitude 0, so I was thinking that maybe the difference in energy relative to the initial situation is the one that was stored in the electric field between them
Sounds good.

lorenz0 said:
, which should be ##\varepsilon_0\int_{R_1}^{R_2}E^2 dV##.
There's a numerical factor missing here. [Edit: @kuruman already noted this.]

lorenz0 said:
This approach would also have the additional benefit of not having to compute the initial and final energy, but does it make sense?
Yes, it does. Good.

lorenz0

## 1. What is the concept of three spherical conductors?

The concept of three spherical conductors refers to a scenario in which three conducting spheres are placed in close proximity to each other, with no physical contact between them. This setup allows for the study of electrical charges and their distribution on the surfaces of the spheres.

## 2. How are the charges distributed on the surfaces of the three spherical conductors?

The charges on the surfaces of the three spherical conductors are distributed in such a way that the electric field inside each conductor is zero. This means that the charges are evenly spread out on the surface, with no net charge inside the conductor.

## 3. What is the significance of studying three spherical conductors?

Studying three spherical conductors can help us understand the behavior of electric charges and fields in a system with multiple conductors. This knowledge can be applied in various fields such as electronics, power transmission, and electrostatics.

## 4. How does the distance between the three spherical conductors affect the distribution of charges?

The closer the three spherical conductors are to each other, the more the charges on their surfaces will interact with each other. This can result in a redistribution of charges, leading to a different electric field configuration. The distance between the conductors also affects the capacitance of the system.

## 5. Can the concept of three spherical conductors be applied in real-life situations?

Yes, the concept of three spherical conductors has practical applications in various fields. For example, it can be used to analyze the behavior of electric charges in a capacitor, which is an essential component in many electronic devices. It can also be applied in the study of lightning and the formation of thunderclouds.

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