Answer to Question #166829 in Electrical Engineering for rishitha

Question

A parallel plate capacitor has an area of 5.00 cm 2 and a capacitance of 3.50 pF. The capacitor is connected to a 12.0 V battery. After the capacitor is completely charged up, the battery is removed a. What must be the separation of the plates, and how much energy is stored in the electric field between the plates? b. If a sheet of Mica with εr=3.5 is placed between the plates, how will the voltage across the plates and the charge density on the plates change? c. If the sheet of Mica has an area of 3.00 cm 2 so that it does not cover the whole area of the plates, what would be the effective capacitance of this arrangement? d. If the Mica sheet is removed and the plates are moved 1.00 mm farther from each other, what will be the voltage between the plates, and how much work will it take to move the plates apart?

Answer 1

a) charge density = $\frac{Q}{Area}$

$= \frac{3.5\times10^-12\times12} {0.0005}$

= $8.4\times 10^-8 C/m^2$

b) charge density = $15.7\times 10^-8 C/m^2$

c) C = 7.35

d) W = $252\times 10^-12$