Question

A charge of $3C$ is moved in an electric field from infinity to two points A and B. If the work done in bringing the charge to point A is $15J$ and in bringing it to point B is $21J$, calculate the potential difference between points A and B.

Answer 1

Step 1: Given data and formula

Given data:

1. Charge is $Q=3C$
2. Work done in moving the charge from infinity to point A is $W_A=15J$
3. Work done in moving the charge from infinity to point B is $W_B=21J$

Formula:

If $W$ is the work done in bringing a charge $\left(Q\right)$ from infinity to a point then the electric potential $\left(V\right)$ at that point is given by

$V=\frac{W}{Q}.........\left(1\right)$

Step 2: Finding the electric potential at point A

Assume $V_A$ be the electric potential at point A then according to the equation $\left(1\right)$

$V_A=\frac{W_A}{Q}.........\left(2\right)$

Substituting the values of $W_A$ and $Q$ into equation $\left(2\right)$ to find the electric potential at point A.

$$\begin{gathered} V_A=\frac{15J}{3C} \\ {V}_A=5V \end{gathered}$$

Step 3: Finding the electric potential at point B

Assume $V_B$ be the electric potential at point B then according to the equation $\left(1\right)$

$V_B=\frac{W_B}{Q}........\left(3\right)$

Substituting the values of $W_B$ and $Q$ into equation $\left(3\right)$ to find the potential at point B.

$$\begin{gathered} V_B=\frac{21J}{3C} \\ {V}_B=7V \end{gathered}$$

Step 4: Calculating the potential difference between points A and B

The potential difference $V^{\text{'}}$between points A and B is the difference of potential at point A and potential at point B. It is given by

$V^{\text{'}}=V_B-V_A........\left(4\right)$

Substituting the values of $V_A$ and $V_B$ obtained from steps 2 and 3 into the equation $\left(4\right)$ to calculate the potential difference.

$$\begin{gathered} V^{\text{'}}=7V-5V \\ {V}^{\text{'}}=2V \end{gathered}$$

Thus, the potential difference between points A and B is $2V$.