Pictured at right is a parallel plate capacitor C, made up of two square plates
ID: 2236769 • Letter: P
Question
Pictured at right is a parallel plate capacitor C, made up of two square plates of side length L separated by a distance d, where d << L. A battery of potential difference Vb is attached along with a switch and a resistor, R, to the capacitor. The switch is initially set to position 1, which charges the capacitor until it is fully charged.
a) Using the definition of capacitance (C = Q/?V), derive the capacitance of our capacitor in terms of d, L, other constants. (Hint: Determine how much charge is on the capacitor when it is fully charged.)
b) Once the capacitor is fully charged, the switch is moved to the neutral position shown above and a rectangular slab of dielectric of dielectric constant ?e, area L x L, and thickness d/4 is inserted between the plates of the conductor. Find the new capacitance of the capacitor. (Hint: Did the charge on the plates change? Did the electric potential difference change?)
Explanation / Answer
a)
C = epsilon0 L^2/d
b)
C' = epsilon0 L^2/(3d/4) + ke epsilon0 L^2/(d/4)
C' = (4 epsilon0 L^2/d) * (1/3 + ke)