# Energy of a charge in between two perpendicular conducting plate .

1. Jun 7, 2013

### bs vasanth

1. Two semi infinite conducting plates at right angles to each other, we bring in a charge q from infinity to point (a,b) what is the energy involved in doing so?

2. Relevant equations
coulomb's law and method of images in griffith.

3. The attempt at a solution
force can be easily calculated by method of images which turns out to:
F=kq^2 [ 1/(2b)^2 - 1/(2a)^2 + 1/((2b)^2+(2a)^2)^0.5]
I can do line integral to this but before that I have to convert them to one single variable, how can i do that ?

2. Jun 7, 2013

### DimReg

I think there's an easier way to do this than doing the line integral. How would you compute this answer if there were no conduction plates, just 4 charges in space (it's the same problem, but I think you're trying too hard because it's a more advanced situation)?

Also, I don't think you've taken into account the directional nature of force in your attempt to solve the problem

3. Jun 8, 2013

### bs vasanth

Yes later I did take directions for the force and then for the upper right charge I resolved the force on the two axis with an unknown angle theta. Now that i have force as :
F= Kq^2[ (-1/(2x)^2 + cos/(2y)^2+(2x)^2 ) i + ( -1/(2y)^2+sin/(2y)^2+(2x)^2 ) j ]
then I did a line integral as for the first term i need to integrate wrt only x and the second term wrt only y .
Other way was finding the potential for the four charges then find the gradient ( field ) and then use the field to find the work done(energy) . But the potential as cross terms and I find difficult to find the gradient. Kindly suggest me if there is a better way.

4. Jun 8, 2013

### DimReg

The answer is simple and it's against forum rules to give the full answer, so sorry if I'm a bit vague.

The electric force is conservative, so the path doesn't matter. This gives a nice expression for the path integrals when the charges are point particles and zero potential is set at infinity. This can be used to compute both work and potential. So if you compute potential then take the gradient and line integral, you're actually going in a circle.

Once you work out the image charges, this is a high school physics problem. Don't over think it.

5. Jun 8, 2013

### bs vasanth

Ok I can just use the potentials to calculate the work done by :W=q.V
W = q.k[ -q/2a -q/2b + q/((2a)^2 + (2b)^2 )^0.5 ]
but should I multiply a negative sign to above equation because those were induced charges in moving q and were not present before , therefore there was negative work done.