Q154 / 160 · TS EAMCET 2020 (09 Sep Shift 2) · Physics
Consider a parallel plate capacitor which is maintained at potential $200 \mathrm{~V}$. The separation distance between the plates of capacitor and the area of plates are $1 \mathrm{~mm}$ and $20 \mathrm{~cm}^2$, respectively. Calculate displacement current in $1 \mu \mathrm{s}$.
- Correct answerA$3.5 \mathrm{~mA}$
- B$2.5 \mathrm{~mA}$
- C$1.5 \mathrm{~mA}$
- D$0.5 \mathrm{~mA}$
Solution
Displacement current,
$\begin{aligned} I_d & =\varepsilon_0 \frac{d \phi}{d t}=\varepsilon_0 \frac{\Delta \phi}{\Delta t} \\ & =\varepsilon_0 \frac{\Delta E A}{\Delta t} \quad(\because \phi=E \cdot A)\end{aligned}$
$=\frac{\varepsilon_0 \cdot \Delta V \cdot A}{\Delta t \cdot d} \quad\left(\because E=\frac{V}{d}\right)$
Here, $\Delta V=200 \mathrm{~V}$,
$\begin{aligned} \Delta t & =1 \mu \mathrm{s}=10^{-6} \mathrm{~s}, \\ d & =1 \mathrm{~mm}=1 \times 10^{-3} \mathrm{~m} \\ \text { and } \quad A & =20 \mathrm{~cm}^2=20 \times 10^{-4} \mathrm{~m}^2\end{aligned}$
$\begin{aligned} \therefore \quad I_d & =\frac{8.85 \times 10^{-12} \times 200 \times 20 \times 10^{-4}}{10^{-6} \times 10^{-3}} \\ & =3.5 \times 10^{-3} \mathrm{~A}=3.5 \mathrm{~mA}\end{aligned}$
Hence, the correct answer is option: 'A'.