4) Calculation Question Answer on a new sheet of paper A parallel plate capacito

Question

4) Calculation Question Answer on a new sheet of paper A parallel plate capacitor has plate area 1 m2 and plate separation of 0.25 cm. Glass with a dielectric constant K = 5 completely fills the space between the conducting plates. The capacitor is fully charged to 12 Volts, then removed from the voltage source. a) What minimum amount of work is required to pull the glass from between the conducting plates? b) After the glass is removed, how much work is required to double the plate separation distance?

Explanation / Answer

here,

area = 1 m^2

d = 0.25 cm = 2.5 * 10^-3 m

the capacitance , C = area * e0/d = 3.54 * 10^-9 F

k = 5 , V = 12 V

when glass is inserted

C' = k * C = 12.5 * 10^-9 F

a)

the minimum amount of work done , W = 0.5 * V^2 * ( C' - C)

W = 0.5 * 12^2 * ( 4 * 2.5 * 10^-9) J

W = 7.2 * 10^-7 J

b)

when the sepration is doubled,

then new capacitance , C1 = C/2

the amount of work done , W = 0.5 * V^2 * ( C - C1)

W = 0.5 * 12^2 * ( 2.5 /2 * 10^-9) J

W = 9 * 10^-8 J

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