A charge of 8 mC is located at the origin.


A charge of $8 \mathrm{mC}$ is located at the origin. Calculate the work done in taking a small charge of $-2 \times 10^{-9} \mathrm{C}$ from a point $\mathrm{P}(0,0,3 \mathrm{~cm})$ to a point $Q(0,4 \mathrm{~cm}, 0)$, via a point $R(0,6 \mathrm{~cm}, 9 \mathrm{~cm})$.


Charge located at the origin, $q=8 \mathrm{mC}=8 \times 10^{-3} \mathrm{C}$

Magnitude of a small charge, which is taken from a point P to point R to point Q, q1 = − 2 × 10−9 C

All the points are represented in the given figure.

Point P is at a distance, d1 = 3 cm, from the origin along z-axis.

Point Q is at a distance, d2 = 4 cm, from the origin along y-axis.

Potential at point $P, V_{1}=\frac{q}{4 \pi \in_{0} \times d_{1}}$

Potential at point $Q, V_{2}=\frac{q}{4 \pi \in_{0} d_{2}}$

Work done (W) by the electrostatic force is independent of the path.

$\therefore W=q_{1}\left[V_{2}-V_{1}\right]$

$=q_{1}\left[\frac{q}{4 \pi \in_{0} d_{2}}-\frac{q}{4 \pi \in_{0} d_{1}}\right]$

$=\frac{q q_{1}}{4 \pi \epsilon_{0}}\left[\frac{1}{d_{2}}-\frac{1}{d_{1}}\right]$  ...(i)

Where, $\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-2}$

$\therefore W=9 \times 10^{9} \times 8 \times 10^{-3} \times\left(-2 \times 10^{-9}\right)\left[\frac{1}{0.04}-\frac{1}{0.03}\right]$

$=-144 \times 10^{-3} \times\left(\frac{-25}{3}\right)$

$=1.27 \mathrm{~J}$

Therefore, work done during the process is 1.27 J.




Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now