# Point charge and conducting sphere

• Tanya Sharma
In summary, the conversation discusses a problem involving a charge placed at a distance from the center of a conducting sphere, and the resulting charge flown through the switch (from ground) when it is closed. The solution to the problem involves using the Method of Image Charges, where the earthed conducting sphere is replaced by an imaginary point charge of the same magnitude as the induced charge on the sphere. The potential at the center of the sphere is calculated to be zero, making it easier to determine the net potential across the volume of the sphere. The conversation also mentions the use of symmetry and the Laplace equation to find the solution.
Tanya Sharma

## Homework Statement

Q a) A charge is placed at a distance x from the center of a conducting sphere of radius R.Find the charge flown through the switch(from ground) when it is closed .

b) In the above question replace the switch with an ammeter .What is the reading of ammeter if charge is at distance 2R from sphere and charge is moved with velocity v towards sphere ?

## The Attempt at a Solution

The charge Q will induce charges on the sphere .Before the switch is closed the potential at the sphere is entirely due to the charge Q,i.e positive potential at the sphere .When the switch is closed the potential on the sphere changes from positive to zero.For that net charge on the sphere must change from initial value zero.

But how will we calculate the potential by charge Q at the sphere .Which point should we use because all points within the sphere are equipotential ? Any point on the surface is as good a candidate as the center or any other point. We may use center of the sphere as its distance from the charge is given .But why not some other point ?

I would be grateful if somebody could help me with the problem.

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• sphere.GIF
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hmm. first, you need to think of how you will solve this problem. there is one external charge, and then when the switch is closed, charges will flow into the sphere, and some kind of equilibrium will be achieved. Keep in mind the charges in the sphere will not necessarily be uniformly distributed in the sphere once equilibrium is achieved (that's a big hint). So how do you calculate the equilibrium when there is one external charge, and some (generally non-uniform) charge distribution inside the sphere?

Is there an alternative method to approach this problem ?

I didn't quite understand the Method of Image Charges :shy:. All I could comprehend was that the earthed conducting sphere could be replaced by an imaginary point charge of same magnitude as that of the induced charge on the sphere .The electric field and potential due to the image charge will be equivalent to that of the sphere .

The solution to the above problem part a) has been given as -

Let the induced charge be q'

Since the sphere is earthed, potential at the center of the sphere is zero .

i.e Kq/x+kq'/R = 0 or q'=-qR/x

But why are they equating potential at the center to be zero ? Why not some other point inside or on the surface of the sphere as it is equipotential ?

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Tanya Sharma said:
But why are they equating potential at the center to be zero ? Why not some other point inside or on the surface of the sphere as it is equipotential ?

Because it is easier to do so. You don't know about the charge distribution of the induced charge but the net induced charge is at equal distance from the centre of sphere so the potential due to them at centre can be easily calculated.

Pranav-Arora said:
Because it is easier to do so. You don't know about the charge distribution of the induced charge but the net induced charge is at equal distance from the centre of sphere so the potential due to them at centre can be easily calculated.

I understand that it is easier to calculate at the center.

But isn't the entire volume of sphere equipotential , having constant potential of kq'/R ?

Tanya Sharma said:
...having constant potential of kq'/R ?

It is at zero potential. :)

1 person
Right...

So if I choose some other point ,then both the contributing terms to the potential change such that the net potential across the volume of the sphere is zero ?

Tanya Sharma said:
So if I choose some other point ,then both the contributing terms to the potential change such that the net potential across the volume of the sphere is zero ?

Yes and it is easiest to do so at the centre.

1 person
Thank you very much.

kq/x+kq'/R = 0 is indeed satisfied, and I can see that the first term represents the potential from q at the centre of the sphere. But what does kq'/R represent ?

Pranav-Arora said:
Because it is easier to do so. You don't know about the charge distribution of the induced charge but the net induced charge is at equal distance from the centre of sphere so the potential due to them at centre can be easily calculated.
What does this mean? at equal distance as what else ?

Story on p 15 UIUC is a little longer. Leads (with a lot of work) to same induced charge, but first determines where the image charge is located.

I must admit I find kq/x+kq'/R = 0 a lot more comfortable, but I can't figure out why it should be valid.

Image charges method point is: For the Laplace equation (ΔV=0), solution outside charges is unique. So if with placing image charges in smart places you can create the same symmetry, etc. as with the conductors with induced charges, you have THE solution.

Here you create an equipotential sphere with radius R by placing a smaller mirror charge on the symmetry axis. There are two unknowns (amount and position) and if sphere is uncharged, intersection points with the symmetry axis give two reasonable equations to find them.

In case of two charges, Q and -q, there is a sphere where the potential is zero.

The opposite is also true. If you have a grounded conducting sphere of radius R and a charge Q outside at distance D from the centre of the sphere, the potential is zero on the sphere and the electric field outside the sphere is the same as if there was a -q charge at the position shown, at distance d=R2/D from the centre of the sphere and q=RQ/D.

In case the sphere is isolated, the sphere is neutral. There is induced negative charge distributed on the outer surface and equal positive charge distributed on the inner surface: The effect of this inside charge distribution is the same as if a positive charge q was at the centre of the sphere. The electric field and potential outside the sphere is the same as that of these three charges: Q, the mirror charge -q and the central charge q.

It s explained for example in Landau-Lifshitz Electrodynamics of Continuous Media.

ehild

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• spheremirror.JPG
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1 person
the method of image of charges is a nice way to solve the problem. But it is not the simplest, I think. For the method of image of charges, you must decide where to place the imaginary charge and what it's magnitude must be, to give an equipotential surface that coincides with the sphere. And then, you have to verify that the magnitude of the imaginary charge is equal to the true total charge which is induced on the surface of the sphere. Because generally, the two things will not have the same value in the method of image of charges.

A much easier method is what (I think) Pranav is describing. You already know the total potential inside the sphere is zero. So you pick the easiest point (the centre) and calculate the potential due to the external charge, and the potential due to the charges on the surface of the sphere. keep working in this general direction, and you can get the answer fairly quickly.

edit: remember, the charge distribution on the sphere is some unknown function. But you don't actually have to calculate it explicitly to find the answer to the question.

Here is the image charge method for a charge outside a conducting circle.

It is quite simple to show that in case of two charges of opposite sign there is a sphere at zero potential. The -q charge is in the origin, and the Q charge is at ρ distance apart on the x axis.
We want to find the points P(x,y) where the potential is equal to zero.

$$U=k\left(\frac{Q}{\sqrt{(x-\rho)^2+y^2}}+\frac{-q}{\sqrt{(x^2+y^2}}\right)=0$$

$$\frac{Q}{\sqrt{(x-\rho)^2+y^2}}=\frac{q}{\sqrt{(x^2+y^2}}$$

Squared and rearranged, we get the equation of a circle:

$$\left(x+\frac{\rho q^2}{Q^2-q^2}\right)^2+y^2=\frac{\rho ^2 Q^2q^2}{(Q^2-q^2)^2}$$

The centre of the circle is at distance ##d= \frac{\rho q^2}{Q^2-q^2}## from the charge q and at distance ##D=\rho+d = \frac{\rho Q^2}{Q^2-q^2}## from the charge Q.

The radius is $$R= \frac{\rho Qq}{|Q^2-q^2|}$$

Substituting ρ = D-d, it follows that

$$\frac{R}{D}=\frac{q}{Q}$$

$$R^2=d D$$

So it is proved that there is a sphere with zero potential in the field of two charges of opposite signs (Q and -q). According to the Uniqueness Theorem if we have a sphere of radius R with zero potential, and a charge Q at distance D from its centre, the electric field is the same as that of the charges Q and -q.

-------------------------------------------------------

Accepting that the method of image charge, the position and magnitude of that virtual charge is easy to obtain: The potential is zero at the closest and farthest points on the sphere.

$$\frac{-q}{R-d}+\frac{Q}{D-R}=0$$

$$\frac{-q}{R+d}+\frac{Q}{D+R}=0$$

From these, R/D=q/Q and R2=dD immediately follows.

ehild

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• twocharges.JPG
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1 person
BvU said:
kq/x+kq'/R = 0 is indeed satisfied, and I can see that the first term represents the potential from q at the centre of the sphere. But what does kq'/R represent ?

What does this mean? at equal distance as what else ?

q' is the induced charge on the surface of the sphere. The elementary surface charge dq' contributes to the potential at the centre by k dq'/R. As every point of the surface is at the same distance from the centre, the total potential of the induced charge is k/R∫dq'=kq'/R.

ehild

ehild said:
... Accepting that the method of image charge, the position and magnitude of that virtual charge is easy to obtain: The potential is zero at the closest and farthest points on the sphere.
that's a nice derivation, thanks. The method of image of charges is almost 'beautiful' in how it can relate two different situations. But we still need to show that the magnitude of the virtual charge is equal to the induced charge on the surface of the sphere in the real configuration. (since that is what the question was asking for). Also, since this is a homework question, maybe we shouldn't give too much away to the OP'er :s

BruceW said:
that's a nice derivation, thanks. The method of image of charges is almost 'beautiful' in how it can relate two different situations. But we still need to show that the magnitude of the virtual charge is equal to the induced charge on the surface of the sphere in the real configuration.

It has been shown by Pranav's method that the induced charge is q'=-QR/D, exactly the same as the image charge.

ehild

right, true. It's a shame the method of image of charges doesn't also give the induced charge in the 'true' configuration. Although I guess since it gives us the potential just outside the sphere, and we know the true potential inside the sphere is zero, then we can calculate the change in potential as we pass from outside to inside the sphere, and therefore get the charge distribution of the true configuration.

BruceW said:
right, true. It's a shame the method of image of charges doesn't also give the induced charge in the 'true' configuration. Although I guess since it gives us the potential just outside the sphere, and we know the true potential inside the sphere is zero, then we can calculate the change in potential as we pass from outside to inside the sphere, and therefore get the charge distribution of the true configuration.

You can determine the electric field E on the outer surface of the sphere using Coulomb's Law, with Q and its image. The surface charge density is ε0E. Note that the electric field can be calculated by the image charge only outside the sphere.

ehild

How can we calculate the answer for the b part ??

Riya ojha said:
How can we calculate the answer for the b part ??
You will need to show your own attempt before help can be given.

## 1. What is a point charge?

A point charge is an idealized model of an electric charge that has no physical size or shape, and is considered to be located at a single point in space. It is used to simplify calculations and understand the behavior of electric charges.

## 2. What is a conducting sphere?

A conducting sphere is a spherical object made of a material that allows electric charges to flow freely through it. This means that the electric charges on the surface of the sphere can move around easily, but the overall charge of the sphere remains constant.

## 3. How does a point charge interact with a conducting sphere?

A point charge will induce an equal and opposite charge on the surface of a conducting sphere. This is known as electrostatic induction. The induced charges will redistribute themselves on the surface of the sphere, resulting in a net electric field inside the sphere that is zero.

## 4. What happens to a point charge inside a conducting sphere?

If a point charge is placed inside a conducting sphere, it will experience a force due to the induced charges on the surface of the sphere. This force will push the point charge towards the surface of the sphere, and it will eventually come to rest at the center of the sphere.

## 5. How can the electric field inside a conducting sphere be calculated?

The electric field inside a conducting sphere can be calculated using Gauss's law. This law states that the electric flux through a closed surface is equal to the net charge enclosed by the surface divided by the permittivity of free space. In the case of a conducting sphere, the net charge enclosed is zero, so the electric field inside the sphere is also zero.

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