Two parallel plate capacitors C1 and C2 are connected in parallel and then to a

Question

Two parallel plate capacitors C1 and C2 are connected in parallel and then to a 24.0 volt battery. Each Plate has area 80.0 cm^2 and the plate separation is 3.00mm. Capacitor C1 is filled with air but capacitor C2 is filled with dielectric constant of k=2.60

a) find capacitance of each capacitor

b) find the charge stored in each capacitor

c) find the total energy stored in the capacitors

The battery is disconnected and then the dielectric is removed from capacitor C2.

d)find the new charge stored in each capacitor

e) find the new voltage across each capacitor

Explanation / Answer

the formula for capacitance for a parallel plate capacitance is keA/d
where k = relative permittivity of the dielectric material between the plates.
and e = permittivity of space
C1 = eA/d = {(8.854 * 10^-12) * (80 * 10^-4)}/ {3 * 10^-3} = 23.6106 * 10^-12 F
C1 = 23.6106 pico Farad
now C2 = k*C1
C2 = 61.3877 pico Farad
___________________________________________________

b)
chage Q = CV
for C1
Q1 = V * C1 = 24 * 23.6106 * 10 ^ -12 = 0.566656 nano coulombs
for C2
Q2 = kQ1 = 1.47331 nano coulombs
___________________________________________________

c) enegry stored = 1/2 cV^2
here V is same for both because of parallel connection
therefore total energy = 1/2 CV^2 { 1 + k} = 24.479 nano Joules
_________________________________________________________

here the voltage supplied is removed and then the dielectric
so here a certain amount of work is done in removing the dielectric
so here we cannot perform the conservation of energy
but we can use the principle of conservation of charge
initially charges are Q1 and Q2
finally both will have same Voltage and same Capacitance
so finally both have same charges
Q1 + Q2 = 2Q
so 20.39616 nano coloumbs = 2Q

d)
so charge on each capacitor = Q = 1.019808 nano coloumbs
_____________________________________________________

e)
new voltage on each capacitor = Q/C = 1.019808/ 23.6106 kV = 43.2 V
______________________________________________________
f)
total energy stored in both the capacitors
1/2cv^2 * 2
= cv^2
=23.6106 pico Farad * 43.2^2 = 44.063 nano joules
_____________________________________________________________

from C) we say that total energy before removal of dielectric
is 24.479 nano Joules
and from f) we say that final total energy is 44.063 nano joules
from this it is cleared that energy is increased because work is done on the system
by some external force in removing dielectric

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