Two identical capacitors (C-1.4 ?F) are connected in parallel, and this combinat

Question

Two identical capacitors (C-1.4 ?F) are connected in parallel, and this combination put in series with a resistor (R 60 k2) and a battery (4V). The capacitors are initially uncharged. 60 k? 1.4 pF a) What is the potential difference across each capacitor 90 ms after the circuit is switched b) A thin copper sheet of thickness 0.1 mm is placed in the (0.5 mm) gap of one of the c) If this new capacitor is used in the circuit and allowed to charge fully, how much on? capacitors, how does its capacitance change? electrical energy is stored in each of the two capacitors?

Explanation / Answer

the two capacitors are connected in parallel.

so, Ceq = 2*C

a) Time constant of the circuit, T = R*Ceq

= 60*10^3*2*1.4*10^-6

= 0.168 s or 168 ms

voltage across capacitor at time t = 90 ms,

V = Vmax*(1 - e^(-t/T))

= 4*(1 - e^(-90/168))

= 1.66 V

b) C' = A*epsilon/d'

= A*epsilon/d)*(d/d')

= C*d/d'

= C*0.5/0.4

= 1.4*0.5/0.4

= 1.75 micro F

c) energy stored in C = 1.4 micr F, U = (1/2)*C*V^2

= (1/2)*1.4*10^-6*4^2

= 1.12*10^-5 J

energy stored in C' = 1.75 micr F, U' = (1/2)*C'*V^2

= (1/2)*1.75*10^-6*4^2

= 1.40*10^-5 J

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