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Q.

Two charges 7 $μ$ C and – 4 $μ$ C are placed at (– 7 cm, 0, 0) and (7 cm, 0, 0) respectively. Given, $\in$ ₀ = 8.85 × 10^–12 C² N^–1 m^–2, the electrostatic potential energy of the charge configuration is :

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a

– 2.0 J

b

– 1.8 J

c

– 1.5 J

d

– 1.2 J

answer is D.

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Detailed Solution

P.E. of two charges
$r = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}}$
= 14 cm

$U = \frac{1}{4 {πε}_{0}} \frac{q_{1} q_{2}}{r}$
$= \frac{9 \times 10^{9} \times 7 \times 10^{- 6} \times (- 4) \times 10^{- 6}}{14 \times 10^{- 2}}$

= – 1.8 J

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Two charges 7 μC and – 4 μC are placed at (– 7 cm, 0, 0) and (7 cm, 0, 0) respectively. Given, ∈0 = 8.85 × 10–12 C2 N–1 m–2, the electrostatic potential energy of the charge configuration is :