Two identical capacitors of area A= 0.002m 2 , andseparation of 1.20 mm are conn

Question

Two identical capacitors of area A= 0.002m2 , andseparation of 1.20 mm are connected in parallel and then connectedto a 12.0 volts battery.
a) Find the capacitance of each capacitor.
b) find the charge on the plates on the capacitors
c) find the total energy stored in the capacitors.

The battery is then removed and the entire space between the platesof one of the capacitors if filled with a dielectric of dielectricconstant k=3 find

d) the new charge on each plate
e) the new voltage across each capacitors
f) find the new total energy stored in the capacitors
g) are the answers to c, and f the same? if not where does theenergy go? or where is it coming from?

Explanation / Answer

C = ε*A/d =  (8.85* 10^−12 *.002)/.00122 = 14.5* 10^-12,
net C of circuit = 2*(14.5 *10^12) = 29*10^-12,
b) charge, Q = C*V = 29*10^-12*12 = 348.19 * 10^-12,,
c) energy, E = .5*C*V^2  = .5 *14.5* 10^-12*12^2 = 10.44 * 10^-10 on each plate.

d) when dielectric is inserted capacitance becomes C = K*ε*A/d so find new capacitance of on capacitor, so there will be potential difference in circuit as Q = CV so  charge in order to become equal will flow and net charge on plates will get decrease, find energy, E = .5*C*V^2 using new found C for one plate and same C for older plate,

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