Question

In order to establish an instantaneous displacement current of 1 mA in the space between the plates of 2 F parallel plate capacitor, the rate of change of potential difference is

• A. $100Vs$
• B. $200Vs$
• C. $300Vs$
• D. $500Vs$

Answer 1

Answer: (D)

$500Vs$

Given : Displacement current $=I_d=1mA=1\times {10}^{-3}A$

$C=2F$

To find : Rate of change of potential difference $=\frac{dc}{dt}=?$

Formula : $I_d={\in }_0\frac{d{\varphi }_E}{dt}$

Sol: Using formula

$I_d={\in }_0\frac{d\varphi }{dt}$

we know, $\varphi =E.A$

where, $E=$ Electric field, $A=Area$

$\therefore I_d={\in }_o.A\frac{dE}{dt}$ ...(1)

We know, $E=\frac{Q}{r^2},v=\frac{Q}{r}$

$\therefore E=\frac{vr}{r^2}=\frac{v}{r}=\frac{v}{d}$ $(\therefore r=d)$

Put above values in eq (1)

$I_d={\in }_0.A\frac{d\left(v/d\right)}{dt}$

here, $d$ is a constant

$\therefore I_d=\frac{{\in }_0.A}{d}\frac{dv}{dt}$

we know, $\frac{{\in }_{0}A}{d}=C=$ capocitance

$\therefore I_d=C.\frac{dv}{dt}$

$\therefore \frac{dv}{dt}=\frac{I_d}{c}=\frac{1\times {10}^{-3}}{2}$

$=0.5\times {10}^{-3}$v/s