- #1

chingel

- 307

- 23

$$E=\frac{xQ^2}{2S\epsilon\epsilon_0}.$$

I would think that the work done goes only into the electrostatic energy between all the charges. There is a charge

$$-Q\left( 1 - \frac{1}{\epsilon} \right)$$

on the dielectric surface right next to the charge ##Q## (and opposite sign charges on the other side, on the plate and dielectric).

When I calculate the total potential energy by summing up over all charges (factor 1/2 is because we have to include each pair only once),

$$E = \frac{1}{2} \sum_{i,j} \frac{k q_iq_j}{r_{ij}},$$

then the contribution from the charge on the dielectric surface cancels out the contribution for an equivalent charge on the plate next to it, leaving a charge

$$Q_2= Q -Q\left( 1 - \frac{1}{\epsilon} \right) = \frac{Q}{\epsilon}$$

on both plates with a remaining contribution. So it seems to be like a capacitor with this new amount of charge on the plates (with no dielectric), which would have total energy

$$E_2 = \frac{xQ^2}{2S\epsilon^2\epsilon_0}.$$

However, this is different from the known answer. What is wrong with this reasoning?