Question

A parallel plate condenser consists of two circular plates each of radius $2cm$ separated by a distance of $0.1mm$. A time varying potential difference of $5\times {10}^{13}v/s$ is applied across the plates of the condenser. The displacement current is:

• A. $5.50A$
• B. $5.56\times {10}^{2}A$
• C. $5.56\times {10}^{3}A$
• D. $2.28\times {10}^{4}A$

Answer 1

Answer: (C)

$5.56\times {10}^{3}A$

Given: A parallel plate condenser consists of two circular plates each of radius 2cm separated by a distance of 0.1mm. A time varying potential difference of $5\times {10}^{13}v/s$ is applied across the plates of the condenser.

To find the displacement current

Solution:

Radius of each circular plate, $r=2cm=0.02m$

Distance between the plates, $d=0.1mm=0.1\times {10}^{-3}m$

the change in potential difference between the plates, $\frac{dV}{dt}=5\times {10}^{13}vs$

Permittivity of free space, ${\epsilon }_0=8.85\times {10}^{-12}{C}^2{N}^{-1}{m}^{-2}$

Hence area of the each plate, $A=\pi r^2=\pi (0.02{)}^2=1.26\times {10}^{-3}{m}^2......(i)$

Capacitance between the two plates is given by the relation,

$C=\frac{{\epsilon }_{0}A}{d}⟹C=\frac{8.85\times {10}^{-12}\times 1.26\times {10}^{-3}}{0.1\times {10}^{-3}}⟹C=11.15\times {10}^{-11}F$

Now we know

displacement current,

$I_d=\frac{dq}{dt}$

where $q$ is the charge on each plate and it is equal to $q=CV$

$I_d=C\frac{dV}{dt}$

By substituting the values, we get

$I_d=11.15\times {10}^{-11}\times 5\times {10}^{13}⟹I_d=57.55\times {10}^2⟹I_d=5.7\times {10}^{3}A$

is the displacement current.