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मूल बिंदु पर एक $8 \,mC$ का आवेश अवस्थित है। $-2 \times 10^{-9}\, C$ के एक छोटे से आवेश को बिंदु $P (0,0,3\, cm )$ से, बिंदु $R (0,6\, cm , 9\, cm )$ से होकर, बिंदु $Q (0,4 \,cm , 0)$ तक ले जाने में किया गया कार्य परिकलित कीजिए
$4.74$
$1.27$
$6.24$
$9.61$
Solution
Charge located at the origin, $q=8 \,mC =8 \times 10^{-3} \,C$
Magnitude of a small charge, which is taken from a point $P$ to point $R$ to point $Q, \;\;q_{1}=-2 \times 10^{-9} \,C$
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 $P, \quad V_{1}=\frac{q}{4 \pi \epsilon_{0} \times d_{1}}$
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. $\therefore W=q_{1}\left[V_{2}-V_{1}\right]$
$=q_{1}\left[\frac{q}{4 \pi \epsilon_{0} d_{2}}-\frac{q}{4 \pi \epsilon_{0} d_{1}}\right]$
$=\frac{q q_{1}}{4 \pi \epsilon_{0}}\left[\frac{1}{d_{2}}-\frac{1}{d_{1}}\right]$
Where, $\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \,N\, m ^{2} \,C ^{-2}$
$\therefore W=9 \times 10^{9} \times 8 \times 10^{-3} \times\left(-2 \times 10^{-9}\right)\left[\frac{1}{0.04}-\frac{1}{0.03}\right]$
$=-144 \times 10^{-3} \times\left(\frac{-25}{3}\right)$
$=1.27 \,J$
Therefore, work done during the process is $1.27 \;J$
