Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsPotential & potential energyMedium2 minPYQ_2021

PhysicsMediumsingle choice

The two thin coaxial rings, each of radius $a$ and having charges $+ Q$ and $- Q$ respectively are separated by a distance of $s$ . The potential difference between the centres of the two rings is :

Options:

Answer:

Solution:

$V_{1} = \frac{K Q}{a} + \frac{- K Q}{\sqrt{s^{2} + a^{2}}}$

$V_{2} = \frac{K Q}{\sqrt{s^{2} + a^{2}}} + \frac{- K Q}{a}$

$V_{1} - V_{2} = \frac{Q}{2 π ε_{0}} [\frac{1}{a} - \frac{1}{\sqrt{s^{2} + a^{2}}}]$

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

⏱ 2mℹ️ Source: PYQ_2021

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