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

Answer:

$Q = -1.43\times 10^[-5} coulomb$

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

$-\frac{Qq}{4\pi \epsilon r^2} = \frac{mv^2}{r}$

solving for Q WE GET

$Q = -\frac{mv^2}{r} \times r^2 \frac{4\pi \epsilon}{q}$

$Q = -mv^2\times r \frac{4\pi \epsilon}{q}$

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

where$\frac{1}{4\pi \epsilon} = 9\times 10^9$

$Q = -1.43\times 10^[-5} coulomb$