Question

An air-filled parallel-plate capacitor has a capacitance of 2.0 F when the plate spacing is 1.6 mm. (a) What is the area of the plates? (b) What is the maximum voltage Vmax that can be applied to this capacitor (before dielectric breakdown occurs)? (c) How much charge is stored on the capacitor when Vmax is across it? (d) How much energy is stored on the capacitor when Vmax is across it? (e) A piece of Plexiglas (with a dielectric constant of 2.1) is inserted between the plates, completely filling the space between the plates, while the capacitor remains connected to a battery of emf equal to Vmax. What are the final, new capacitance, charge, and energy?

Answer 1

Answer a)

Now,

where, C = capacitance = 2.0

d = seperation between the plates = 1.6mm

and = 8.54 × 10−12 F/m

Now , A =[ ( 2*10^-6 F) * (1.6 * 10^-3 m) ] / 8.54 × 10−12 F/m

A = 374.707m2 .

Answer b)  

Electric potential difference across cap. plates and E field inside related by separation d at breakdown at critical field

= 4800 V

Answer c)

Now,

Qmax = 9.6 mC

Answer d)

Energy Stored =

U = 23.04 J

Answer e)

when dielectric material is inserted between parallel plates the capacitance increases and the new capacitance C' is given by C' = k×C, where k is dielectric constant.

New Capacitance = 4.2 micro F

Charge Q remains same. Hence potential difference reduces and the new value of potential difference is V/k .

new Vbr = (4800/2.1) = 2285.71 V

Energy stored in the capacitor before inserting dielectric = (1/2)×C×V2

after inserting dielectric, energy available in the capacitor = (1/2)×(kC)×(V/k)2 = k×(1/2)×C×V2

Energy Stored = 2.1 * U = 2.1 * 23.04 = 48.384 J