मूल बिंदु पर एक 8 ,mC का आवेश अवस्थित है। -2 | vedclass

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Question

Charge located at the origin, $q=8 \,mC =8 	imes 10^{-3} \,C$

Magnitude of a small charge, which is taken from a point $P$ to point $R$ to point $

Vedclass · Accepted Answer

$1.27$

Vedclass · Answer

Charge located at the origin, $q=8 \,mC =8 	imes 10^{-3} \,C$

Magnitude of a small charge, which is taken from a point $P$ to point $R$ to point $Q, \;\;q_{1}=-2 	imes 10^{-9} \,C$

All the points are represented in the given figure.

Point $P$ is at a distance, $d_{1}=3 \,cm ,$ from the origin along $z$ -axis. Point $Q$ is at a distance, $d _{2}=4 \,cm ,$ from the origin along $y$ -axis.

Potential at point $P, \quad V_{1}=\frac{q}{4 \pi \epsilon_{0} 	imes d_{1}}$

Potential at point $Q$, $\quad V_{2}=\frac{q}{4 \pi \epsilon_{0} d_{2}}$

Work done $(W)$ by the electrostatic force is independent of the path. $	herefore W=q_{1}\left[V_{2}-V_{1}ight]$

$=q_{1}\left[\frac{q}{4 \pi \epsilon_{0} d_{2}}-\frac{q}{4 \pi \epsilon_{0} d_{1}}ight]$

$=\frac{q q_{1}}{4 \pi \epsilon_{0}}\left[\frac{1}{d_{2}}-\frac{1}{d_{1}}ight]$

Where, $\frac{1}{4 \pi \epsilon_{0}}=9 	imes 10^{9} \,N\, m ^{2} \,C ^{-2}$

$	herefore W=9 	imes 10^{9} 	imes 8 	imes 10^{-3} 	imes\left(-2 	imes 10^{-9}ight)\left[\frac{1}{0.04}-\frac{1}{0.03}ight]$

$=-144 	imes 10^{-3} 	imes\left(\frac{-25}{3}ight)$

$=1.27 \,J$

Therefore, work done during the process is $1.27 \;J$

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