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PhysicsElectrostaticsPotential & potential energyMedium2 minQB

PhysicsMediuminteger

Two charges $+ 2 q$ and $- 3 q$ are fixed at $(4 m, 0, 0)$ and $(9 m, 0, 0)$ respectively. The distance between 2 points on $X$ -axis (in $m$ ) where the potential is zero is $x$ . Then the value of $\frac{x}{2}$ is (Assume standard reference point for electric potential)

Answer:

Solution:

Potential at P \& at Q is zero

$\Rightarrow \frac{2 q}{x} = \frac{3 q}{5 - x} \Rightarrow x = 2 m & \frac{2 q}{y} = \frac{3 q}{5 + y} \Rightarrow y = 10 m \Rightarrow PQ = x + y = 12 m$

Stream:JEESubject:PhysicsTopic:ElectrostaticsSubtopic:Potential & potential energy

⏱ 2mℹ️ Source: QB

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