In order to establish an instantaneous displacement current of $$1\,mA$$ in the space between the plates of $$2\mu F$$ parallel plate capacitor, the potential difference need to apply is
A.
$$100\,V{s^{ - 1}}$$
B.
$$200\,V{s^{ - 1}}$$
C.
$$300\,V{s^{ - 1}}$$
D.
$$500\,V{s^{ - 1}}$$
Answer :
$$500\,V{s^{ - 1}}$$
Solution :
$$\eqalign{
& {I_d} = 1\,mA = {10^{ - 3}}A \cr
& C = 2\mu F = 2 \times {10^{ - 6}}F \cr
& {I_D} = {I_C} = \frac{d}{{dt}}\left( {CV} \right) = C\frac{{dV}}{{dt}} \cr} $$
Therefore, $$\frac{{dV}}{{dt}} = \frac{{{I_D}}}{C} = \frac{{{{10}^{ - 3}}}}{{2 \times {{10}^{ - 6}}}} = 500\,V{s^{ - 1}}$$
Therefore, applying a varying potential difference of $$500\,V{s^{ - 1}}$$ would produce a displacement current of desired value.
Releted MCQ Question on Electrostatics and Magnetism >> Electromagnetic Waves
Releted Question 1
Out of the following options which one can be used to produce a propagating electromagnetic wave?
The electric field associated with an electromagnetic wave in vacuum is given by $$E = \hat i40\cos \left( {kz - 6 \times {{10}^8}t} \right),$$ where $$E,z$$ and $$t$$ are in $$volt/m,$$ metre and second respectively. The value of wave vector $$k$$ is