Question

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 ______.

Options

• `(q_0Q)/(4piε_0a)`

• `(q_0Q)/(8piε_0a)`

• `(q_0Q)/(4piε_0a^2)`

• `(q_0Q)/(8piε_0a^2)`

Answer 1

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`