An infinite plane sheet has charge density. obtain the expression for the work done in bringing a point charge q from infinity to a point,distant r in front of the charged plane sheet.
Answers
First the set up. Let's assume I have an infinite charged plate with some constant charge density over the plate, say σσ. That means that I have σCm2σCm2 over the plate where C is Coulomb and m is meters.
I did the math and found that the electric field at any point is 2πKσ2πKσ where KK is Coulomb's constant. That means the Force at any given point doesn't depend upon the distance from the plate and we get Fe=2πKσqFe=2πKσq for some other particle with charge qq.
Now, I learned that Electric Potential is equal to KQrKQr. I tried to derive this and I think it comes from taking the Force formula F=KQqd2F=KQqd2 dividing by qq to get a "per unit charge" and then integrating out from ∞∞ to rr. Basically I integrate out the work per charge to move the particle from an infinite distance to r away from the particle with charge Q.
However, I don't think that the formula works universally. If I have that same infinite plate, then F=2πKσqF=2πKσq. Doing the a related calculation for work on some charge coming in from infinity to r is:
W=−∫r∞Fds=−∫r∞2πKσqds=−2πKσq∫r∞ds=∞W=−∫∞rFds=−∫∞r2πKσqds=−2πKσq∫∞rds=∞.
Since the work is ∞∞ that means the electric potential is infinite? That seems to me to be intuitively right. I need to add up a finite fixed amount infinitely many times as I move the charged particle in. As long as the finite fixed amount is some ϵ>0ϵ>0, that would be infinity.
But that means the electric potential is infinite which is a direct contradiction to the formula KQr<∞KQr<∞. So does that formula no longer hold for a plate? It's only for a point charge somewhere in space?