Two charges 7 and –4 are placed at (–7 cm, 0, 0) and (7 cm, 0, 0) respectively. Given , the electrostatic potential energy of the charge configuration is: [2025]

Question

Two charges 7 $μ C$ and –4 $μ C$ are placed at (–7 cm, 0, 0) and (7 cm, 0, 0) respectively. Given $ε_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ , the electrostatic potential energy of the charge configuration is: [2025]

Smart Achievers · Accepted Answer

(4)

P.E. of two charges, $U = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r}$

$r = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} {+ (z_{2} - z_{1})}^{2}} = 14 cm$

$\Rightarrow U = \frac{9 \times 10^{9} \times 7 \times 10^{- 6} \times (- 4) \times 10^{- 6}}{14 \times 10^{- 2}}$ = –1.8 J

Solution

Q.
Two charges 7 $μ C$ and –4 $μ C$ are placed at (–7 cm, 0, 0) and (7 cm, 0, 0) respectively. Given $ε_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ , the electrostatic potential energy of the charge configuration is: [2025]

1 –1.5 J

2 –2.0 J

3 –1.2 J

4 –1.8 J

Ans.
(4)

P.E. of two charges, $U = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r}$

$r = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} {+ (z_{2} - z_{1})}^{2}} = 14 cm$

$\Rightarrow U = \frac{9 \times 10^{9} \times 7 \times 10^{- 6} \times (- 4) \times 10^{- 6}}{14 \times 10^{- 2}}$ = –1.8 J

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