# Griffiths Problem 3.34

1. Mar 10, 2008

### ehrenfest

1. The problem statement, all variables and given/known data
This question refers to Griffiths E and M book.

If they had not told me that the charge moves, I would have guessed that it just sits there since it seems like there is nothing to exert a force on it. Am I supposed to calculate the induced charge on the conducting plate? How do I use the fact that the plate is grounded? That means that the potential on the plate is 0 just like at infinity, but doesn't that imply there is no charge on the plate either?

2. Relevant equations

3. The attempt at a solution

2. Mar 10, 2008

### pam

No, the charge on the plate is needed to keep it at V=0.
The charge is supplied by the ground.

3. Mar 10, 2008

### ehrenfest

A related question: on page 123, Griffiths says "Evidently the total induced charge on the plane is -q, as (with the benefit of hindsight) you can perhaps convince yourself that it had to be."

I cannot convince myself unfortunately. How do you know this a priori? Is it related to the fact that the plate is grounded? Would it be different if the plate were not grounded?

4. Mar 10, 2008

### ehrenfest

And also when I try to apply eqn 3.12 to this problem I get a differential equation which I have no idea how to solve:

$$\ddot{z} = C/z^2$$

where C is a constant. What am I doing wrong?

5. Mar 10, 2008

### pam

1. The image charge is -q. By Gauss's law, that has to equal the induced surface charge on the grounded plane.

2. Use dv/dt=v(dv/dx).

6. Mar 11, 2008

### Reshma

Calculate the force on the charge at a height 'x' above the plane and equate it with the equation of motion of the charge 'q'.

7. Mar 11, 2008