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Q.

The electric field between the plates of a parallel-plate capacitor of capacitance $2 .0 μF$ drops to one third of its initial value in $4 .4 μs$ when the plates are connected by a thin wire. Find the resistance of the wire. ( Given ln3 = 1.1)

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a

2Ω

b

3Ω

c

4Ω

d

5Ω

answer is B.

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Detailed Solution

The electric field between the plates is
$E = \frac{Q}{{Aε}_{0}} = \frac{Q_{0}}{{Aε}_{0}} e^{- t / RC} Or E = E_{0} e^{- t / RC}$
In the given problem, $E = \frac{1}{3} E_{0} at t = 4 .4 μs$
Thus $\frac{1}{3} = e^{- \frac{4 .4 μ s}{RC}} Or, \frac{4 .4 μs}{RC} = \ln 3 = 1 .1 Or, R = \frac{4 .4 μs}{1 .1 \times 2 .0 μF} = 2 .0 Ω$

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The electric field between the plates of a parallel-plate capacitor of capacitance 2.0μF drops to one third of its initial value in 4.4μs when the plates are connected by a thin wire. Find the resistance of the wire. ( Given ln3 = 1.1)