Question 28: EM waves

Question

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

Accepted Answer

For a capacitor, the displacement current is given by
$$ I_d = C \frac{dV}{dt} $$
Given:
$$ I_d = 4.0 \, 	ext{A} $$
$$ C = 6\,\mu	ext{F} = 6 	imes 10^{-6}\,	ext{F} $$
We need to find:
$$ \frac{dV}{dt} $$
Using the formula,
$$ \frac{dV}{dt} = \frac{I_d}{C} $$
Substitute the values:
$$ \frac{dV}{dt} = \frac{4.0}{6 	imes 10^{-6}} $$
$$ = \frac{2}{3} 	imes 10^6 $$
$$ = 0.67 	imes 10^6 \,	ext{V/s} $$
So,
$$ \alpha = 0.67 $$
Hence, the correct option is
B
.

Answer

0.58

Answer

0.67

Answer

0.82

Answer

0.75

EM waves

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

Choose the correct answer: