JEE Main 26-Aug-2021 Shift 1 Solved Paper

The given situation is shown below Resistivity, ρ = 200 Ωm
C=2‌pF=2×10−12F
V = 40 V
K or εr = 50
ileakage=?
C=

Kε0A

d

(K is the dielectric constant or relative permittivity)
and R=

ρd

A

Now, charge will be discharged through the resistance between theplates.
Now, time constant (τ) of discharging,
τ=RC=

ρd

A

×

Kε0A

d

τ=ρKε0
For a given R-C circuit, the discharged current is given by
i=

Q

RC

e−

t

RC

i=

Q

pKε0

e−

t

pKε0

The above discharge current is the leakage current,
ileakage=

Q

ρKε0

e−

t

ρKε0

Maximum leakage current,
(i0)leakage=

Q

ρKε0

=

CV

ρKε0

=

2×10−1240

200×50×8.85×10−12

= 903 µA = 0.9 mA
∴ (i0)leakage=0.9‌mA