Work done to insert a point charge 𝑞 at the center of a conducting sphere

[songoku]

Homework Statement

How much work is needed to insert a point charge 𝑞 at the center of a conducting charged sphere of radius 𝑅, with a total charge 𝑄 uniformly distributed? Assuming the point charge is brought in from infinity and 𝑞 ≪ 𝑄.

Relevant Equations

$F=\frac{kQq}{r^2} \hat r$
$W=\int \vec F . d\vec s$

Electric field inside conducting charged sphere is zero so the potential inside it will be constant, hence there will be no work to move a charge from the surface to the center. It means the work done is for moving the point charge 𝑞 to the surface of the conducting charged sphere.

The $d\vec s$ is $\hat r~dr$ but I am not sure about the vector form of the force. How to know whether the force is $\frac{kQq}{r^2} \hat r$ or $\frac{kQq}{r^2} (-\hat r)$

Thanks

 

[TSny]

You want the work you do while moving $q$. What direction is your force?

 

[kuruman]

The $d\vec s$ is $\hat r~dr$ but I am not sure about the vector form of the force. How to know whether the force is $\frac{kQq}{r^2} \hat r$ or $\frac{kQq}{r^2} (-\hat r)$

In that regard, the question is ambiguous. Not only the sign of the charge is not specified but also the force that is doing this work. Is it the electrical force or the agent that moves the charge?

Just find the absolute value of the work needed to bring the charge from infinity to the surface. What assumption must you make given that $q<<Q$ ?

 

[songoku]

In that regard, the question is ambiguous. Not only the sign of the charge is not specified but also the force that is doing this work. Is it the electrical force or the agent that moves the charge?

I will just assume both charges are positive and the work done is by external force.

What assumption must you make given that $q<<Q$ ?

I am sorry I don't know the assumption needed. I don't think I make any assumptions in my working.

You want the work you do while moving $q$. What direction is your force?

It would be radially inward because I want to bring it from infinity to the surface. So:
$$W=\int^{R}_{\infty} \left(\frac{kQq}{r^2}(-\hat r)\right).(\hat r~dr)$$
$$=\frac{kQq}{r}$$

If the question is asking about the work done by electric field, then the answer would be $-\frac{kQq}{r}$

Is this correct? Thanks

 

[TSny]

I will just assume both charges are positive and the work done is by external force.


I am sorry I don't know the assumption needed. I don't think I make any assumptions in my working.


It would be radially inward because I want to bring it from infinity to the surface. So:
$$W=\int^{R}_{\infty} \left(\frac{kQq}{r^2}(-\hat r)\right).(\hat r~dr)$$
$$=\frac{kQq}{r}$$

Looks good except your answer is expressed in terms of $r$ instead of $R$.

You don't need to make any assumptions about the signs of the charges. Your answer will take care of all possibilities.

 

[Steve4Physics]

@songoku - just in case you are not already aware - by using the potential at the sphere's surface you can answer the question without integration.

 

[vela]

I am sorry I don't know the assumption needed. I don't think I make any assumptions in my working.

In your calculation, your expression for the force relies on the assumption that the distribution of charge on the sphere remains spherically symmetric, but because the sphere is conductive, the presence of $q$ will induce polarization. With the assumption that $Q \gg q$, you can neglect the effect of polarization.

 

[jbriggs444]

In your calculation, your expression for the force relies on the assumption that the distribution of charge on the sphere remains spherically symmetric, but because the sphere is conductive, the presence of $q$ will induce polarization. With the assumption that $Q \gg q$, you can neglect the effect of polarization.

The distribution of charge $Q$ will be symmetric at the initial state with infinite separation. And it will be symmetric in the end state with the travelling charge centered.

As long as we are considering only electrostatics and a perfectly conducting sphere, what happens in between will not affect the required work. All that matters is the potentials of the end states. We need not consider the permutations of charge that might occur during the transit.

If the conducting sphere had resistance, that could add to the work required. If we were considering electrodynamics then some EM radiation could be produced, adding to the work required.

 

[PhDeezNutz]

Oh my goodness, please don’t do a line integral. As stated above that would assume the charges on the sphere don’t move around while the charges come close.

Make use of the potential energy formulation.

 

[vela]

The distribution of charge $Q$ will be symmetric at the initial state with infinite separation. And it will be symmetric in the end state with the travelling charge centered.

As long as we are considering only electrostatics and a perfectly conducting sphere, what happens in between will not affect the required work. All that matters is the potentials of the end states. We need not consider the permutations of charge that might occur during the transit.

Yup, but that's a different argument/calculation than what @songoku did. His calculation, which is likely the one expected for this problem, relies on the assumption that $q \ll Q$.

 

[TSny]

Using the method of images, it appears that the work to bring $q$ to the surface of the conducting sphere diverges no matter how small the value of $q$. According to Jackson's text (page 59 in 2nd edition), the force on $q$ as a function of $r$ is $$\mathbf F = \frac {kq} {r^2}\left[Q - \frac{qR^3(2r^2-R^2)}{r(r^2-R^2)^2} \right] \hat{r} $$

 

[songoku]

@songoku - just in case you are not already aware - by using the potential at the sphere's surface you can answer the question without integration.

Do you mean like this?
$$W=q.\Delta V$$
$$=q.(V_{\text{at surface}}-V_{\infty})$$
$$=q.\left(\frac{kQ}{R}-0\right)$$
$$=\frac{kQq}{R}$$

 

[songoku]

Thank you very much for all the help and explanation TSny, kuruman, Steve4Physics, vela, jbriggs444, PhDeezNutz