Where should $q_{3}$ be placed to make the 'potential energy of the system equal to zero ?

X = 0.743 m

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X = 0.0743 m

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X = 0.00743 m

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X = 0

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Solution

The correct option is B $X = 0.0743 m$

Total PE of system

$\frac{1}{4 π ϵ} \frac{q_{1} q_{2}}{r_{12}} + \frac{1}{4 π ϵ} \frac{q_{3} q_{2}}{r_{23}} + \frac{1}{4 π ϵ} \frac{q_{1} q_{3}}{r_{13}} = 0$

$\Rightarrow \frac{- 12}{0.2} + \frac{- 6}{0.2 - X} + \frac{8}{X} = 0$

$X = 0.074266 m O R 0.359 m$

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Where should q3 be placed to make the 'potential energy of the system equal to zero ?

The correct option is B X=0.0743m Total PE of system 14πϵq1q2r12+14πϵq3q2r23+14πϵq1q3r13=0 ⇒−120.2+−60.2−X+8X=0 X=0.074266m OR 0.359m

Where should $q_{3}$ be placed to make the 'potential energy of the system equal to zero ?