# Potential energy due to an external charge and a grounded sphere

• phantomvommand
In summary: Rd/(d^2 - R^2)^2, or something else entirely. The limits of integration are not affected by this.In summary, the force between the point charge and the sphere is given by Coulomb's law, which can be expanded to take into account the potential energy between the point charge and the image charge. It is possible to redefine the axes such that the origin is at the image charge, but the force between the point charge and the sphere will remain the same.
phantomvommand
Homework Statement
*The sphere is metal, so charges effectively lie on the surface.
In terms of the external charge q, radius of the sphere R and the distance d (distance between external charge and centre of sphere)
determine the following electrostatic energies:
a) the electrostatic energy of the interaction between charge q and the induced
charges on the sphere;
b) the electrostatic energy of the interaction among the induced charges on the
sphere;
c) the total electrostatic energy of the interaction in the system.
Relevant Equations
Method of image charges gives the following:
The metal sphere can be replaced with a charge q' = -q* R/d,
at a distance
d' = R^2/d from the centre of the sphere.
Energy can be found by integrating -Fdx.

Let us attempt part C first, which is to find the total energy of the entire system.
I can definitely find an expression for the force, as given by Coulomb's Law. However, why should I integrate this force from infinity to d, where d is the distance of the external charge to the centre of the sphere?

I was thinking that the upper limit of the integration should be d - d', which is the distance between the external charge and the image charge. OR, If you consider the limiting case where R --> infinity, the external charge is effectively facing a flat plate, and so the upper limit of integration should be d - R. Why is this wrong?

For part (a), the electrostatic energy between the external charge and the induced charges is apparently just the energy between the external charge and the image charge, as given by kqq'/(d-d'). Why is this true? Aren't the formulas for energy inapplicable for regions within the sphere, as the image charge only replicates conditions outside the sphere?

Any help would be greatly appreciated. Thank you!

Last edited:
phantomvommand said:
Let us attempt part C first, which is to find the total energy of the entire system.
I can definitely find an expression for the force
I assume you mean the force between the point charge and the sphere, or, equivalently, between the point charge and the image charge.
But be careful here: as you integrate wrt movement of the point charge, are you keeping the image charge constant or allowing that to move and weaken in response to the movement of the point charge?

Last edited:
phantomvommand said:
I was thinking that the upper limit of the integration should be d - d', which is the distance between the external charge and the image charge.
The upper limit will depend on the choice of the variable that you use to express the force. If you choose the origin at the center of the sphere and let ##x## be the distance of the charge ##q## from the origin, then you can express the force on ##q## in terms of ##x##. You can then find the work for moving ##q## from infinity to the final position by letting the limits of integration be infinity and the final value of ##x##. How is the final value of ##x## related to ##d## as defined in the problem?

Delta2
TSny said:
The upper limit will depend on the choice of the variable that you use to express the force. If you choose the origin at the center of the sphere and let ##x## be the distance of the charge ##q## from the origin, then you can express the force on ##q## in terms of ##x##. You can then find the work for moving ##q## from infinity to the final position by letting the limits of integration be infinity and the final value of ##x##. How is the final value of ##x## related to ##d## as defined in the problem?

the force is kqq’/(d-d’)^2, as given by Coulomb’s law. This can be expanded out to be kq^2Rd/(d^2 - R^2)^2. I suppose in this case, when integrating wrt “x”, “x” is taken with origin at the centre, since the original formula for the force has the length d, which is the distance of the external charge to the centre of the sphere.
However, why can’t I simply redefine the axes, such that the origin is now at the image charge? Wouldn’t the expression for the force remain the same, except now I am integrating from infinity to a distance d - d’, thus giving a different result? Thank you.

haruspex said:
I assume you mean the force between the point charge and the sphere, or, equivalently, between the point charge and the image charge.
But be careful here: as you integrate wrt movement of the point charge, are you keeping the image charge constant or allowing that to move and weaken in response to the movement of the point charge?

I remember that the point charge isn’t allowed to move, otherwise you would be getting energy “for free”. I suppose it is kept constant then.

phantomvommand said:
the force is kqq’/(d-d’)^2, as given by Coulomb’s law. This can be expanded out to be kq^2Rd/(d^2 - R^2)^2. I suppose in this case, when integrating wrt “x”, “x” is taken with origin at the centre, since the original formula for the force has the length d, which is the distance of the external charge to the centre of the sphere.
Yes. Both ##x## and ##d## are locating the charge ##q## relative to the center of the sphere. Think of ##d## as representing the particular value of ##x## for which you want to calculate the potential energies (as shown in the figure). So, for an arbitrary value of ##x##, the force may be expressed as a function of ##x## as ##F(x) =kq^2Rx/(x^2 - R^2)^2##. What would be the value of ##x## for the upper limit of integration when finding the work done by ##F(x)## in going from infinity to the position of ##q## shown in the figure?

However, why can’t I simply redefine the axes, such that the origin is now at the image charge? Wouldn’t the expression for the force remain the same, except now I am integrating from infinity to a distance d - d’, thus giving a different result? Thank you.
I'm not following how you are going to do this. As you move the charge ##q## from infinity to the position shown in the figure, the position of the image charge ##q'## will also move. Are you taking your origin to move with the movement of ##q'##, or are you taking your origin to be fixed at the specific location of ##q'## as shown in the figure? Either way, can you show explicitly how you would set up the integral for either of these choices?

Delta2
phantomvommand said:
the point charge isn’t allowed to move
Eh? Imagining it moving to infinity is how you calculate the PE.
As you quoted,
The metal sphere can be replaced with a charge q' = -q* R/d, at a distance d' = R^2/d from the centre of the sphere.
If you actually move the point charge to infinity, integrating the force as you go, the point will also move, and change in charge. What will the force be when the point charge is at distance x from the sphere's centre?

haruspex said:
Eh? Imagining it moving to infinity is how you calculate the PE.
As you quoted,
The metal sphere can be replaced with a charge q' = -q* R/d, at a distance d' = R^2/d from the centre of the sphere.
If you actually move the point charge to infinity, integrating the force as you go, the point will also move, and change in charge. What will the force be when the point charge is at distance x from the sphere's centre?
Thank you for this. This idea of image charges is very new to me, so I was not aware that the image charge also moves. I now see that because the point charge also moves, it is important that we define the force with respect to a fixed displacement from a fixed axis, which can be taken to be the centre of the sphere. The point charge is not fixed, and thus cannot be taken as a fixed axis. Thus, using the formulas for the charge and distance of the image charge, a force integral with constant terms, apart from the distance to centre of sphere, can be set up.

TSny said:
Yes. Both ##x## and ##d## are locating the charge ##q## relative to the center of the sphere. Think of ##d## as representing the particular value of ##x## for which you want to calculate the potential energies (as shown in the figure). So, for an arbitrary value of ##x##, the force may be expressed as a function of ##x## as ##F(x) =kq^2Rx/(x^2 - R^2)^2##. What would be the value of ##x## for the upper limit of integration when finding the work done by ##F(x)## in going from infinity to the position of ##q## shown in the figure?

I'm not following how you are going to do this. As you move the charge ##q## from infinity to the position shown in the figure, the position of the image charge ##q'## will also move. Are you taking your origin to move with the movement of ##q'##, or are you taking your origin to be fixed at the specific location of ##q'## as shown in the figure? Either way, can you show explicitly how you would set up the integral for either of these choices?
Thank you for this. I was not aware that the point charge also moves. I now see the importance of substituting in the values of the image charge and its distance from the centre of sphere; this is to set up an integral that can be integrated with respect to a fixed axis (centre of sphere). Integrating with respect to distance to the moving point charge is pointless, as the distance and force constantly changes.

## 1. What is potential energy due to an external charge and a grounded sphere?

Potential energy due to an external charge and a grounded sphere is the energy stored in the system when a charged object is placed near a grounded conducting sphere. This energy is a result of the interaction between the external charge and the electric field created by the grounded sphere.

## 2. How is potential energy calculated in this system?

The potential energy in this system can be calculated using the formula U = (Q1Q2)/4πεr, where Q1 is the external charge, Q2 is the charge of the grounded sphere, ε is the permittivity of the medium, and r is the distance between the two charges.

## 3. What happens to the potential energy if the external charge is moved closer to the grounded sphere?

If the external charge is moved closer to the grounded sphere, the potential energy will increase. This is because the electric field between the two charges will become stronger, resulting in a greater interaction and therefore a higher potential energy.

## 4. Can the potential energy in this system ever be negative?

No, the potential energy in this system can never be negative. This is because the charges in the system have the same sign and therefore the potential energy will always be positive. Negative potential energy would only occur if the charges had opposite signs.

## 5. How does the potential energy change if the grounded sphere is replaced with an insulating sphere?

If the grounded sphere is replaced with an insulating sphere, the potential energy in the system will decrease. This is because the electric field created by the insulating sphere will not be as strong as that of a grounded sphere, resulting in a weaker interaction and therefore a lower potential energy.

Replies
1
Views
645
Replies
23
Views
1K
Replies
4
Views
834
Replies
21
Views
1K
Replies
2
Views
505
Replies
6
Views
985
Replies
28
Views
2K
Replies
22
Views
2K
Replies
18
Views
2K
Replies
23
Views
836