Question

A source of charge $Q$ is kept at a distance of $3km$ from a test charge ${q_0}$ . If $Q$ is displaced slightly, then the change in the electric field detected by ${q_0}$ will be after:
(A) $10s$
(B) $10ms$
(C) $10\mu s$
(D) $10ns$

Answer 1

If the source charge is displaced at a distance from a test charge, then the change in electric field detected by the test charge will be the time taken to complete the given displaced distance. Here, Ampere’s law will be applied.

Complete answer:
Distance of the source charge from a test charge, $d = 3km = 3 \times {10^3}m$
If the source charge $Q$ displaces, then the change in the electric field will be the time taken by source charge to displace at the test charge:
$\therefore time = \dfrac{{dis\tan ce}}{{speed}}$
Constant speed is $3 \times {10^8}m.{s^{ - 1}}$
$\Rightarrow t = \dfrac{{3 \times {{10}^3}}}{{3 \times {{10}^8}}}$
$\Rightarrow t = {10^{ - 5}}\sec$
or $t = 10 \times {10^6}\sec = 10\mu s$
Therefore, the time taken is the change in electric field will be $10\mu s$ .

Hence, the correct option is (C) $10\mu s$ .

Note:
The rate of change of the electric charge stored on the plates of a parallel plate capacitor is equal to the product of the permittivity of free space, the area of the capacitor's plate, and the rate of change of the electric field between the plates, according to Ampere's Law in its general version.