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प्रश्न
A point charge q0 is moving along a circular path of radius a, with a point charge Q at the centre of the circle. The kinetic energy of q0 is ______.
विकल्प
`(q_0Q)/(4piε_0a)`
`(q_0Q)/(8piε_0a)`
`(q_0Q)/(4piε_0a^2)`
`(q_0Q)/(8piε_0a^2)`
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उत्तर
A point charge q0 is moving along a circular path of radius a, with a point charge Q at the centre of the circle. The kinetic energy of q0 is `underlinebb((q_0Q)/(8piε_0a))`.
Explanation:
q0 is moving around a circular path of radius a with a point charge Q at the centre of the circle. To calculate the kinetic energy of q0, we can put,
Electrostatic force due to charges = Centripetal force
`(kQq_0)/r^2 = (mv^2)/r`
The radius of the circle given in the question is a
`(kQq_0)/a^2 = (mv^2)/a`
`(kQq_0)/a = mv^2` ............(i)
Kinetic energy is equal to `1/2m"v"^2`.
Multiplying (i) by `1/2`,
`1/2(kQq_0)/a = 1/2mv^2`
`1/2 xx 1/((4piepsilon_0)) xx Q xx q_0/a = 1/2mv^2`
`(Qq_0)/(8piepsilon_0a) = 1/2mv^2`
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