Question

If the work done by the electric field to bring a point charge of $5\mu C$ from a point A to a point B is $10{\text{mJ}}$ , then find the potential difference ${V_A} - {V_B}$ .

A) +2 kV

B) −2 kV

C) +2000V

D) −2000V

Answer 1

The work done per charge will correspond to the potential difference between the two points. Charge moves from one point to another only when there is a potential difference between the two points and it always moves from a point of high potential to a point of low potential.

Formula used:

-The potential difference between the points A and B is given by, ${V_A} - {V_B} = \dfrac{W}{q}$ ,

where ${V_A}$ and ${V_B}$ are the electrostatic potentials at point A and point B respectively, $W$ is the work done and $q$ is the amount of the point charge.

Complete step by step answer.

Step 1: List the parameters involved in the problem at hand.

Here, a point charge of $q = 5\mu C$ is brought from a point A to a point B. The work done in bringing this charge is given as $W = 10{\text{mJ}}$ .

Let ${V_A}$ and ${V_B}$ be the electrostatic potentials at the points A and B respectively.

Step 2: Express the relation for the potential difference between point A and point B.

The potential difference between the points A and B is given by, ${V_A} - {V_B} = \dfrac{W}{q}$ --- (1)

where ${V_A}$ and ${V_B}$ are the electrostatic potentials at point A and point B respectively, $W$ is the work done and $q$ is the amount of the point charge.

Step 3: Substitute the values for $q$ and $W$ in equation (1) to find the potential difference between the points ${V_A} - {V_B}$ .

The amount of the point charge is $q = 5\mu C$ and the work done is $W = 10{\text{mJ}}$

${V_A} - {V_B} = \dfrac{{10 \times {{10}^{ - 3}}}}{{5 \times {{10}^{ - 6}}}} = + 2000{\text{V}}$

∴ The required potential difference ${V_A} - {V_B} = 2000{\text{V}} = 2{\text{kV}}$ .

Therefore, both the options, A) and C) are correct.

Note: When substituting values in an equation make sure that all the involved physical quantities are expressed in their respective S.I. units. Necessary conversions must be done if this is not the case. Here, the charge moves from point A (higher potential) to point B (lower potential). Thus the required potential difference ${V_A} - {V_B}$ is positive.