The electric field between the two parallel plates of a capacitor of 1.5 $μ F$ capacitance drops to one third of its initial value in 6.6 $μ s$ when the plates are connected by a thin wire.

The resistance of this wire is __________ $Ω .$ (Given, log 3 = 1.1) [2024]

Question

The electric field between the two parallel plates of a capacitor of 1.5 $μ F$ capacitance drops to one third of its initial value in 6.6 $μ s$ when the plates are connected by a thin wire.

The resistance of this wire is __________ $Ω .$ (Given, log 3 = 1.1) [2024]

Smart Achievers · Accepted Answer

(4) $E = \frac{Q}{A ϵ_{0}} \Rightarrow E \propto Q \Rightarrow E = E_{0} e^{- t / R C}$

$\Rightarrow t = R C \ln \frac{E_{0}}{E} \Rightarrow 6.6 \times 10^{- 6} = R (1.5 \times 10^{- 6}) \ln 3$

$\Rightarrow R = \frac{66}{151 \ln 3} = \frac{66}{15 \times 1.1} = 4 Ω$

Solution

Q.
The electric field between the two parallel plates of a capacitor of 1.5 $μ F$ capacitance drops to one third of its initial value in 6.6 $μ s$ when the plates are connected by a thin wire.

The resistance of this wire is __________ $Ω .$ (Given, log 3 = 1.1) [2024]

Ans.
(4) $E = \frac{Q}{A ϵ_{0}} \Rightarrow E \propto Q \Rightarrow E = E_{0} e^{- t / R C}$

$\Rightarrow t = R C \ln \frac{E_{0}}{E} \Rightarrow 6.6 \times 10^{- 6} = R (1.5 \times 10^{- 6}) \ln 3$

$\Rightarrow R = \frac{66}{151 \ln 3} = \frac{66}{15 \times 1.1} = 4 Ω$

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Solution

Q. The electric field between the two parallel plates of a capacitor of 1.5 μF capacitance drops to one third of its initial value in 6.6 μs when the plates are connected by a thin wire. The resistance of this wire is __________ Ω. (Given, log 3 = 1.1) [2024]

Ans. (4) E=QAϵ0⇒E∝Q⇒E=E0e-t/RC ⇒t=RClnE0E⇒6.6×10-6=R(1.5×10-6)ln3 ⇒R=66151ln3=6615×1.1=4Ω

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Q.
The electric field between the two parallel plates of a capacitor of 1.5 $μ F$ capacitance drops to one third of its initial value in 6.6 $μ s$ when the plates are connected by a thin wire.

The resistance of this wire is __________ $Ω .$ (Given, log 3 = 1.1) [2024]

Ans.
(4) $E = \frac{Q}{A ϵ_{0}} \Rightarrow E \propto Q \Rightarrow E = E_{0} e^{- t / R C}$

$\Rightarrow t = R C \ln \frac{E_{0}}{E} \Rightarrow 6.6 \times 10^{- 6} = R (1.5 \times 10^{- 6}) \ln 3$

$\Rightarrow R = \frac{66}{151 \ln 3} = \frac{66}{15 \times 1.1} = 4 Ω$