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NCERT Class XII Chapter
Electrostatic Potential and Capacitance
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Question : 12 of 37
Marks: +1, -0
A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of – 2 × 10−910^{-9} C from a point P(0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm).
Solution:  
As electric field is conservative field, so work done in moving a charge in electric field is independent of path chosen to move the charge in electric field and depends only on the electric potential difference between the two end points. So
WPQW_{PQ} = q0[VQ−VP]q_0[V_Q-V_P] = q0[14πε0qrOQ−14πε0qrOP]q_0\left[\frac{1}{4\pi\varepsilon_0}\frac{q}{r_{OQ}} - \frac{1}{4\pi\varepsilon_0}\frac{q}{r_{OP}}\right]
or WPQW_{PQ} = qq04πε0[1rOQ−1rOP]\frac{qq_0}{4\pi\varepsilon_0}\left[\frac{1}{r_{OQ}} - \frac{1}{r_{OP}}\right]
= 9 × 10910^9 × 8 × 10−310^{-3} C × (−2×10−9 C)(-2\times 10^{-9}\,\mathrm{C}) × (14−13)\left(\frac{1}{4} - \frac{1}{3}\right) × 110−2\frac{1}{10^{-2}}
= - 144 × 10−110^{-1} × (3−412)\left(\frac{3-4}{12}\right) = 12 × 10−110^{-1}
WPQW_{PQ} = 1.2 J
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