Question

A displacement current of 4.0 A can be set up in the space between two parallel plates of $6 \mu \text{F}$ capacitor. The rate of change of potential difference across the plates of the capacitor is nearly $\alpha \times 10^6 \text{ V/s}$. The value of $\alpha$ is _________.

• A. 0.58
• B. 0.67
• C. 0.82
• D. 0.75

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The topic of this question is Electromagnetic Waves, specifically focusing on displacement current.
We are asked to find the rate of change of voltage ($dV/dt$) given the capacitance and the displacement current inside the capacitor.
Step 2: Key Formula or Approach:
The displacement current is given by the formula:
\[ I_d = C \frac{dV}{dt} \]
Step 3: Detailed Explanation:
We are given $I_d = 4.0 \text{ A}$ and $C = 6 \mu \text{F} = 6 \times 10^{-6} \text{ F}$.
Substitute the values into the formula:
\[ 4.0 = (6 \times 10^{-6}) \times \frac{dV}{dt} \]
\[ \frac{dV}{dt} = \frac{4.0}{6 \times 10^{-6}} = \frac{2}{3} \times 10^6 \approx 0.666 \dots \times 10^6 \text{ V/s} \]
Rounding off to two decimal places gives $\alpha = 0.67$.
Step 4: Final Answer:
The value of $\alpha$ is 0.67.