Question

A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.923 g, q = 4.52 µC is located on the x axis at x = 22.6 cm, moving with a speed of 45.7 m/s in the positive y direction. For what value of Q will the moving particle execute circular motion? (Neglect the gravitational force on the particle.)

Answer 1

`Q = -1.43* 10^[-5} coulomb`

Step-by-step explanation:

Given data:

particle mass = 0.923 g

particle charge is 4.52 micro C

speed of particle 45.7 m/s

In this particular case, coulomb attraction will cause centrifugal force and taken as +ve and Q is taken as -ve

`-(Qq)/(4\pi \epsilon r^2) = (mv^2)/(r)`

solving for Q WE GET

`Q = -(mv^2)/(r) * r^2 (4\pi \epsilon)/(q)`

`Q = -mv^2* r (4\pi \epsilon)/(q)`

`Q = - \frac{0.923* 10^(-3) * 45.7^2* (22.6* 10^(-2))} {4.52* 10^(-6) * 9* 10^9}`

where
`(1)/(4\pi \epsilon) = 9* 10^9`

`Q = -1.43* 10^[-5} coulomb`