Question

Calculate the amount of work done in carrying a charge of 5 mC against a potential difference of 100 V.
(A) 0.5 J
(B) 50 J
(C) 5 J
(D) None

Answer 1

Hint
The amount of work done in carrying a charge is given by the product of the amount of charge and the potential difference.

Complete step by step answer
Given, the amount of charge, $\mathrm{q}=5\mathrm{m}\mathrm{C}=5\times {10}^{-3}$ C
The amount of potential difference, $\mathrm{V}=100\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{t}$
The electric potential difference is the amount of work done required to move a unit electric charge from one point to another point.
Thus, the work done on the charge is given by,
$\mathrm{W}=\mathrm{C}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}\times \mathrm{P}\mathrm{o}\mathrm{t}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{D}\mathrm{i}\mathrm{f}\mathrm{f}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}$
$\mathrm{W}=5\times {10}^{-3}\times 100$
$\mathrm{W}=0.5$J
Therefore, (A) $0.5$J is the required solution.

Additional information Electric potential is the amount of work done required to move a charged particle from infinity to a certain point. It is denoted by the letter ‘V’. Its SI unit is ‘Volts’ and is represented by the letter ‘V’. Electric potential is measured by the device ‘Voltmeter’. It should be connected in parallel. Hence, an electric potential is constant in parallel connection.

Note
In the case of a system of two positive charges at a distance, the value of work done computed could be negative if a unit positive charge is moving towards the other positive charge.