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 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 _________.
Updated On: Apr 12, 2026
- 0.58
- 0.67
- 0.82
- 0.75
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
Displacement current $I_d$ is numerically equal to the conduction current $I_c$ flowing into the capacitor plates. We need to find the rate of change of voltage ($dV/dt$) given $I_d$ and capacitance $C$.
Step 2: Key Formula or Approach:
$I_d = \epsilon_0 \frac{d\Phi_E}{dt} = C \frac{dV}{dt}$
Step 3: Detailed Explanation:
Given $I_d = 4.0 \text{ A}$ and $C = 6 \mu \text{F} = 6 \times 10^{-6} \text{ F}$.
The relation is:
\[ I_d = C \frac{dV}{dt} \]
\[ 4.0 = (6 \times 10^{-6}) \times \frac{dV}{dt} \]
\[ \frac{dV}{dt} = \frac{4.0}{6 \times 10^{-6}} = \frac{2}{3} \times 10^6 \approx 0.666 \dots \times 10^6 \text{ V/s} \]
Rounding to two decimal places, we get $\alpha = 0.67$.
Step 4: Final Answer:
The value of $\alpha$ is 0.67.
Displacement current $I_d$ is numerically equal to the conduction current $I_c$ flowing into the capacitor plates. We need to find the rate of change of voltage ($dV/dt$) given $I_d$ and capacitance $C$.
Step 2: Key Formula or Approach:
$I_d = \epsilon_0 \frac{d\Phi_E}{dt} = C \frac{dV}{dt}$
Step 3: Detailed Explanation:
Given $I_d = 4.0 \text{ A}$ and $C = 6 \mu \text{F} = 6 \times 10^{-6} \text{ F}$.
The relation is:
\[ I_d = C \frac{dV}{dt} \]
\[ 4.0 = (6 \times 10^{-6}) \times \frac{dV}{dt} \]
\[ \frac{dV}{dt} = \frac{4.0}{6 \times 10^{-6}} = \frac{2}{3} \times 10^6 \approx 0.666 \dots \times 10^6 \text{ V/s} \]
Rounding to two decimal places, we get $\alpha = 0.67$.
Step 4: Final Answer:
The value of $\alpha$ is 0.67.
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