A leaky parallel plate capacitor is filled completely with a material having dielectric constant K=5 and electrical conductivity σ=7.4×10−12Ω−1 m−1. If the charge on the capacitor at instant t=0 is q=8.85μC, then calculate the leakage current at the instant t=12 s.

Name: Video solution 1: A leaky parallel plate capacitor is filled completely with a material having dielectric constant K=5 and electrical conductivity σ=7.4×10−12Ω−1 m−1. If the charge on the capacitor at instant t=0 is q=8.85μC, then calculate the leakage current at the instant t=12 s.
Uploaded: 2022-07-14T17:21:31.000Z
Duration: 5 min
Description: Video solution 1: A leaky parallel plate capacitor is filled completely with a material having dielectric constant K=5 and electrical conductivity σ=7.4×10−12Ω−1 m−1. If the charge on the capacitor at instant t=0 is q=8.85μC, then calculate the leakage current at the instant t=12 s.

This is basically a problem of discharging of CR circuit, because between the plates of the capacitor, there is capacitor as well as resistance.R=σAd(R=σAl)and C=dKε0A∴ Time constant,τc=CR=σKε0Substituting the values, we haveτc=7.4×10−125×8.86×10−12=5.98 sCharge at any time decreases exponentially asq=q0e−t/τeHere, q0=8.85×10−6C ( Charge at time t=0 )Therefore, discharging (leakage) current at time t will be given byi=(−dtdq)=τcq0e−t/τcor current at t=12 s isii=5.98(8.85×10−6)e−12/5.98=0.198×10−6 A=0.198μA=0.198μA

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Electrostatic Potential and Capacitance

Effect of Dielectric on Capacitance

A leaky parallel plate capacitor is filled completely with a m

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A leaky parallel plate capacitor is filled completely with a material having dielectric constant $K = 5$ and electrical conductivity $σ = 7.4 \times 1 0^{- 12} Ω^{- 1} m^{- 1}$ . If the charge on the capacitor at instant $t = 0$ is $q = 8.85 μ C$ , then calculate the leakage current at the instant $t = 12 s$ .

Views: 5,706 students

Updated on: Jul 14, 2022

Exam:

JEE Advanced 1997

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Owen Discussed

A leaky parallel plate capacitor is filled completely with a material having dielectric constant

K = 5

and electrical conductivity

σ = 7.4 \times 1 0^{- 12} Ω^{- 1} m^{- 1}

. If the charge on the capacitor at instant

t = 0

q = 8.85 μ C

, then calculate the leakage current at the instant

t = 12 s

9 mins ago

Discuss this question LIVE

9 mins ago

Text solutionVerified

This is basically a problem of discharging of

CR

circuit, because between the plates of the capacitor, there is capacitor as well as resistance.

R = \frac{d}{σ A} (R = \frac{l}{σ A})

and

C = \frac{K ε _{0} A}{d}

∴

Time constant,

τ_{c} = CR = \frac{K ε _{0}}{σ}

Substituting the values, we have

τ_{c} = \frac{5 \times 8.86 \times 1 0 ^{- 12}}{7.4 \times 1 0 ^{- 12}} = 5.98 s

Charge at any time decreases exponentially as

q = q_{0} e^{- t / τ_{e}}

Here,

q_{0} = 8.85 \times 1 0^{- 6} C

( Charge at time

t = 0

)
Therefore, discharging (leakage) current at time

t

will be given by

i = (- \frac{d q}{d t}) = \frac{q _{0}}{τ _{c}} e^{- t / τ_{c}}

or current at

t = 12 s

i i = \frac{( 8.85 \times 1 0 ^{- 6} )}{5.98} e^{- 12/5.98} = 0.198 \times 1 0^{- 6} A = 0.198 μ A = 0.198 μ A

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Question Text	A leaky parallel plate capacitor is filled completely with a material having dielectric constant $K = 5$ and electrical conductivity $σ = 7.4 \times 1 0^{- 12} Ω^{- 1} m^{- 1}$ . If the charge on the capacitor at instant $t = 0$ is $q = 8.85 μ C$ , then calculate the leakage current at the instant $t = 12 s$ .
Updated On	Jul 14, 2022
Topic	Electrostatic Potential and Capacitance
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 2
Upvotes	303
Avg. Video Duration	8 min

Text solutionVerified

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A leaky parallel plate capacitor is filled completely with a material having dielectric constant K=5 and electrical conductivity σ=7.4×10−12Ω−1 m−1. If the charge on the capacitor at instant t=0 is q=8.85μC, then calculate the leakage current at the instant t=12 s.

Text solutionVerified

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