# Point charge and conducting sphere

1. Mar 12, 2014

### Tanya Sharma

1. The problem statement, all variables and given/known data

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 ?

2. Relevant equations

3. 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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2. Mar 12, 2014

### BruceW

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?

3. Mar 12, 2014

### ehild

4. Mar 17, 2014

### Tanya Sharma

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 ?

Last edited: Mar 17, 2014
5. Mar 17, 2014

### Saitama

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.

6. Mar 17, 2014

### Tanya Sharma

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 ?

7. Mar 17, 2014

### Saitama

It is at zero potential. :)

8. Mar 17, 2014

### Tanya Sharma

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 ?

9. Mar 17, 2014

### Saitama

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

10. Mar 17, 2014

### Tanya Sharma

Thank you very much.

11. Mar 17, 2014

### BvU

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 ?

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.

12. Mar 18, 2014

### ehild

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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13. Mar 18, 2014

### BruceW

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.

14. Mar 18, 2014

### ehild

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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15. Mar 18, 2014

### ehild

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

16. Mar 18, 2014

### BruceW

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

17. Mar 18, 2014

### ehild

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

ehild

18. Mar 18, 2014

### BruceW

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.

19. Mar 19, 2014

### ehild

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

20. May 6, 2017

### Riya ojha

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