Answer to Question #177679 in Electric Circuits for Salman Wajeeh

Question

A parallel plate capacitor (with air between its plates and with plates whose area can be changed) is charged by a battery until the charge stored is Q . While the capacitor is still connected to the battery, the capacitor's plates are pushed farther apart so the distance between them is multiplied by two and are made smaller so the cross sectional area of the plates is divided by four. A dielectric slab (with dielectric constant 6.0) is inserted between the plates completely filling the space between them. What will be the charge stored after the dielectric is inserted?

Answer 1

Find initial capacitance:

$$C_1=\frac{\epsilon_0 A}{d}.$$

Voltage applied:

$$V=\frac QC=\frac{Qd}{\epsilon_0 A}.$$

Find final capacitance:

$$C_2=\frac{k\epsilon_0A_2}{d_2}=\frac{6\epsilon_0(A/4)}{2d}=\frac{3\epsilon_0A}{4d}.$$

The final charge with the modifications done to the capacitor and dielectric inserted:

$$q=VC_2=\frac{Qd}{\epsilon_0 A}·\frac{3\epsilon_0A}{4d}=\frac34 Q.$$