Question

5) Three capacitors, all initially without any dielectrics inserted, of C1 = 3 micro-farads, C2 = 7 micro-farads, and C3 = 15 micro-farads, are connected purely in series to a battery and the total stored energy of all three capacitors is 203 micro-joules. Then, the plates of C1 are brought closer by a factor of 2 and a dielectric with a dielectric constant of 1.7 is inserted into C2 while still connected in series to the same battery. (C3 does not change at all.) How much does the total stored energy of all three capacitors increase in micro-joules after the capacitors are changed? Not the factor change... how much does the stored energy go up in value.

Answer 1

When three capacitors are connected in series, their equivalent capacitance will be:

1/Ceq = 1/C1 + 1/C2 + 1/C3

Ceq = [1/3 + 1/7 + 1/15]^-1 = 1.84 micro-farads

Ceq = 1.84*10^-6 F

Now energy stored in capacitors is given by:

U = (1/2)*Ceq*V^2

V = Voltage of battery = sqrt(2U/Ceq)

Using given values:

V = sqrt (2*203*10^-6/(1.84*10^-6)) = 14.85 V

Now when plates of C1 is brought closer, it's new capacitance will be

C1' = e0*A'/d'

Since A' = A & d' = d/2, then

C1 = e0*A/(d/2) = 2*e0*A/d = 2*C1 = 2*3 = 6 micro-farads

When dielectric is inserted between C2, new capacitance will be

C2' = k*C2 = 1.7*7 = 11.9 micro-farads

there is no change in C3, So

C3' = C3 = 15 micro-farads

Now equivalent capacitance will be

1/Ceq' = (1/C1' + 1/C2' + 1/C3')

Ceq' = [1/6 + 1/11.9 + 1/15]^-1

Ceq' = 3.15 micro-farads = 3.15*10^-6 F

Now energy stored in capacitors will be

U' = (1/2)*Ceq'*V^2

Given that battery is not changed, So V = 14.85 V

So,

U' = (1/2)*3.15*10^-6*14.85^2

U' = 347.3*10^-6 J = 347.3 micro-joules

So energy stored is increased by:

dU = U' - U = 347.3 - 203 = 144.3 micro-joules

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