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Question
A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (U) as ε = αU where α = 2V–1. A similar capacitor with no dielectric is charged to U0 = 78V. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.
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Solution
Let the final voltage be U: If C is the capacitance of the capacitor without the dielectric, then the charge on the capacitor is
Q1 = CU
The capacitor with the dielectric has a capacitance εC. Hence the charge on the capacitor is
Q2 = εU = α CU2
The initial charge on the capacitor that was charged is
Q0 = CU0
From the conservation of charges,
Q0 = Q1 + Q2
Or, CU0 = CU + α CU2
⇒ αU2 + U – u0 = 0
∴ U = `(-1 +- sqrt(1 + 4αU_0))/(2α)`
= `(-1 +- sqrt(1 + 624))/4`
= `(-1 +- sqrt(625))/4` volts
As U is positive
U = `(sqrt(625) - 1)/4 = 24/4` = 6V
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