Question

A capacitor has been charged by a dc source. What are the magnitude of conduction current
and displacement currents?
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Answer 1

Answer:

$\text{\underline{Formulae:}}\\\\1.\: \text{Conduction Current:} = C.\dfrac{dv}{dt}\\\\2.\: \text{Displacement Current:} = C\dfrac{d\phi}{dt}$

Assuming the Capacitor to be fully charged, I am solving this question.

$\text{Electric Flux} (d\phi) = d[\dfrac{q}{\epsilon_0}]\\\\\text{Since q is constant, dq will be constant due to which Flux becomes zero.}$

$\text{Hence Displacement Current is Capacitance multiplied to zero, which is zero.}$

$\text{Similarly Change in Potential for a Fully Charged Capacitor is zero.}$

$\text{Hence Change in Potential becomes zero. Therefore Conduction current in zero.}$

Hope it helped!

Answer 2

Answer:

Explanation:

Refer the below attachment.