An air-filled, parallel-plate capacitor with area A and gap width d is connected

Question

An air-filled, parallel-plate capacitor with area A and gap width d is connected to a battery that maintains the plates at potential difference V. The plates are pulled apart so that the gap is 5 times wider, while they remain in electrical contact with the battery terminals. By what factor does the potential energy of the capacitor charge? (express your answer to two significant figures.) By what factor does the potential energy of the capacitor change when the gap becomes 5 times wider, if the capacitor is removed from the battery and thus isolated? (express your answer to two significant figures.)

Explanation / Answer

C1 = eo*A/d1

C2 = eo*A/d2

d2 = 5d1

C2 = eo*A/5d1 = C1/5

Ui = 0.5*C1*V1^2

after the gap is increases by 5 times

V2 = v1

Uf =.5*C2*v2^2 = 0.5*C1/5*v1^2 = (1/5)*Ui

Uf/Ui = 1/5 <<-----answer

+++++++++++++

Ui = 0.5*Q1^2/C1

Uf = 0.5*Q2^2/C2

C2 = C1/5

Q2 = Q1

Uf = 0.5*Q1^2/(C1/5) = 5*Ui

Uf/Ui = 5 <--------------answer

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