Question

#PHYSICS

Q. Two point charges +q and -q, each of mass m, are revolving in a circle of radius R. Under mutual electrostatic force, what will be the total energy of the system?
(Consider only electrostatic force)

#Explanation is compulsory
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Answer 1

Answer:

System will rotate about axis passing through COM.

Centripetal force is provided by electrostatic attraction.

So: $\boxed{\sf{\frac{1}{4\pi\in_{0}} \frac{q.q}{(2R) ^{2} } = \frac{ {mv}^{2} }{R}}}$

That is: $\boxed{\sf{{mv}^{2} = \frac{ {q}^{2} }{16\pi\in_{0} R}}}$

Now:

$\implies$ ${\sf{(K.e.)_{system} = 2 \times \frac{1}{2} {mv}^{2}}}$

$\implies$ $\sf{mv^{2} = \frac{ {q}^{2} }{16\pi\in_{0} R}}$

And,

Potential energy:

$\implies$ $\sf{ \frac{1}{4\pi\in_{0}R} \frac{(q)( - q)}{2R}}$

$\implies$ $\sf{- \frac{q^{2}}{8\pi\in_{0} R}}$

Total energy of system (K.e.) + P. E.

$\implies$ $\sf{ \frac{ {q}^{2} }{16\pi\in_{0}R} - \frac{ {q}^{2} }{8\pi\in_{0}R}}$

$\implies$ $\sf{ - \frac{ {q}^{2} }{16\pi\in_{0}R}}$

Therefore:

Total energy of the system: $\boxed{\sf{ - \frac{ {q}^{2} }{16\pi\in_{0}R}}}$