A system of three charges are placed as shown in the figure: If D >> d, the potential energy of the system is best given by: (a) (1)/(4πε₀)[q²/d+(qQd)/(D²)] (b) (1)/(4πε₀)[-q²/d-(qQd)/(2D²)] (c) (1)/(4πε₀)[-q²/d-(qQd)/(D²)] (d) (1)/(4πε₀)[-q²/d+(2qQd)/(D²)]

Answer: (c). D is measured from the centre of the +q,-q pair, so +q is at D + d/2 and -q at D - d/2 from Q. U = (1)/(4πε₀)[-q²/d + qQ((1)/(D+d/2) - (1)/(D-d/2))] For D d: (1)/(D+d/2) - (1)/(D-d/2) ≈ -d/D² U = (1)/(4πε₀)[-q²/d - (qQd)/(D²)], answer (c).

A system of three charges are placed as shown in the figure: If D >>… | JEE Main 2019 (9 Apr, Shift 1)

Q49JEE Main 2019 · 9 APR · SHIFT 1Medium

Electrostatic Potential Energy

A system of three charges are placed as shown in the figure:

If D >> d, the potential energy of the system is best given by:

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Answer: (c)

$D$ is measured from the centre of the $+q,-q$ pair, so $+q$ is at $D + \frac{d}{2}$ and $-q$ at $D - \frac{d}{2}$ from $Q$ .
$U = \frac{1}{4\pi\varepsilon_0}\left[-\frac{q^2}{d} + qQ\left(\frac{1}{D+d/2} - \frac{1}{D-d/2}\right)\right]$
For $D \gg d$ : $\frac{1}{D+d/2} - \frac{1}{D-d/2} \approx -\frac{d}{D^2}$
$U = \frac{1}{4\pi\varepsilon_0}\left[-\frac{q^2}{d} - \frac{qQd}{D^2}\right]$ , answer (c).

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Electrostatic Potential Energy question from JEE Main 2019 (9 Apr, Shift 1)

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