Two parallel plates are connected to a battery which maintains a constant potent

Question

Two parallel plates are connected to a battery which maintains a constant potential difference of 18.0 V. The distance between the plates can be varied and there are three plates of different dielectric materials available: rubber (0.20 mm, ? = 6.7), glass (5 mm, ? = 7) and teflon (2 cm, ? = 2.1). The area of the plates is 83 cm2. What is the maximum energy that can be stored in the capacitor using one of the three materials?


Attempt:
v = 18
k = 6.7
d = 0.00002
a = 0.0083

C = k*eps*A/d
C = 2.4590482288676093637774022230344*10^-8

U = .5*C*V^2
U = .5*2.4590482288676093637774022230344*10^-8*18^2
U = 3.9836581307655271693193916013157e-6 J

Explanation / Answer

energy stored in a capacitor U = 1/2 CV2

since ,V is same for all the capacitor . Energy will be maximum for the maximum capacitance.

capacitance C = kA/d

A and are same for all the capacitors . Therefore , the capacitance will be maximum for the maximum value of k/d of the dieletric.

for rubber k/d = 6.7 / 0.20 = 33.5

for glass k/d = 7/5 =1.4

for teflon k/d = 2.1/20 = 0.105 [ 2 cm = 20 mm]

thus , for the rubber the stored energy will be maximum

for rubber C = kA/d = 2.46 x 10-9 F

U = 0.5 * (2.46 x 10-9)* 182 =3.98 x 10-7 J

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