Question

Two particles are fixed on an x axis. Particle 1 of charge 69.9 μC is located at x = -4.32 cm; particle 2 of charge Q is located at x = 36.1 cm. Particle 3 of charge magnitude 22.1 μC is released from rest on the y axis at y = 4.32 cm. What is the value of Q if the initial acceleration of particle 3 is in the positive direction of (a) the x axis and (b) the y axis?

Answer 1

a)
`1.45* 10^(-2)\ \rm C`

b)
`1.76*10^(-3)\ \rm C`

Step-by-step explanation:

Given:

Charge on particle(1) =
`69.9\ \mu C`

Charge on particle(3) =
`22.1\ \mu C`

Angle made by particle 1 with the x axis

`\tan \theta=(4.32)/(4.32)\\\theta =45^\circ`

Angle made by particle 3 with the x axis

`\tan \theta=(4.32)/(36.1)\\\theta =6.82^\circ`

a) If the acceleration particle 3 is along the positive x axis then then it means the net force on the charge along y axis is zero so

`(k* 69.9)/(r_1^2)* \sin45^\circ=(k* Q)/(r_2^2)* \sin6.28^\circ\\(k* 69.9)/(37.32)* \sin45^\circ=(k* Q)/(1321.8)* \sin6.28^\circ\\Q=1.45*10^(-2)\ \rm C`

b) If the acceleration particle 3 is along the positive y axis then then it means the net force on the charge along x axis is zero so

`(k* 69.9)/(r_1^2)* \cos45^\circ=(k* Q)/(r_2^2)* \cos6.28^\circ\\(k* 69.9)/(37.32)* \cos45^\circ=(k* Q)/(1321.8)* \cos6.28^\circ\\\\Q=1.76*10^(-3)\ \rm C`