Question

A parallel plate capacitor is charged to \[60\mu C\].Due to a radioactive source,the plate loses charge at the rate of \[1.8 \times {10^{ - 8}}\,C/s\].The magnitude of displacement current is:
A. \[1.8 \times {10^{ - 8}}C/s\]
B. \[{\text{3}}{\text{.6}} \times {10^{ - 8}}C/s\]
C. \[{\text{4}}{\text{.1}} \times {10^{ - 11}}C/s\]
D. \[{\text{5}}{\text{.7}} \times {10^{ - 12}}C/s\]

Answer 1

Hint: To solve the given question we use the concepts of displacement current. Displacement current is the rate of change of electrical displacement field. Displacement current differs from the conduction current because the displacement current does not involve electrons' movement

Formulae used:
\[{I_d} = \dfrac{{dq}}{{dt}}\]
Where, ${I_d}$ is the displacement current, $q$ is the charge and $t$ is the time.

Complete step by step answer:
The capacitor is charged initially upto \[60\,\mu C\], then it loses charge because of a radioactive source. We need to seek out the displacement current. The displacement current is that current which comes into play within the region during which the electrical field and hence the electric flux is changing with time.

Maxwell found that conduction current (\[I\]) and displacement current (\[{I_d}\]​) together have the property of continuity, although individually they may not be continuous. Maxwell also predicted that this current produces an equivalent magnetic flux as a conduction current can produce. Displacement current is given by,
\[{I_d} = \dfrac{{dq}}{{dt}}\]
As displacement current is the rate of change of electric displacement field. The rate of loss of charge of the capacitor is given as \[1.8 \times {10^{ - 8}}C/s\].Therefore
\[{I_d} = \dfrac{{dq}}{{dt}} \\
\therefore {I_d} = 1.8 \times {10^{ - 8}}C/s\]

Therefore, the correct answer is option \[{\text{A}}\].

Note: The displacement of the present is directly proportional to the voltage within the circuit and inversely proportional to the resistance of the capacitor. As the voltage within the circuit increases, the displacement current within the circuit also increases. As the voltage within the circuit decreases, the displacement current within the circuit also decreases. If the resistance within the circuit increases, the displacement current decreases.If the resistance within the circuit decreases, the displacement current increases.
circuit increases, the displacement current decreases. If the resistance in the circuit decreases, the displacement current increases.