The electric field between the two parallel plates of a capacitor of 1.5 capacitance drops to one third of its initial value in 6.6 when the plates are connected by a thin wire. The resistance of this wire is ________ Ω. (Given, log e 3=1.1) Numerical answer type. Work it out, then open the answer.

Answer: 4. E = (q)/(ε₀ A) ∝ q, and on discharge q = q₀e-t/RC. e-t/RC = 1/3 ⇒ t = RCln 3 R = (t)/(Cln 3) = 6.6×10⁻⁶1.5×10⁻⁶× 1.1 R = 4 Ω

The electric field between the two parallel plates of a capacitor of… | JEE Main 2024 (5 Apr, Shift 1)

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Capacitors in DC Circuits & RC

The electric field between the two parallel plates of a capacitor of $1.5\ \mu\text{F}$ capacitance drops to one third of its initial value in $6.6\ \mu\text{s}$ when the plates are connected by a thin wire. The resistance of this wire is ________ $\Omega$ . (Given, $\log_e 3=1.1$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 4

$E = \frac{q}{\varepsilon_0 A} \propto q$ , and on discharge $q = q_0e^{-t/RC}$ .
$e^{-t/RC} = \frac{1}{3} \Rightarrow t = RC\ln 3$
$R = \frac{t}{C\ln 3} = \frac{6.6\times10^{-6}}{1.5\times10^{-6}\times 1.1}$
$R = 4\ \Omega$

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