Question

Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at y2 = +0.34 m. A third point charge q = +8.4 μC is fixed at the origin. The net electrostatic force exerted on the charge q by the other two charges has a magnitude of 25 N and points in the +y direction. Determine the magnitude of q2.

Answer 1

`50.91 \mu C`

Step-by-step explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for
`q_(1)` and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by
`q_(1)` on q. Then we can know the magnitude of the force exerted by
`q_(2)` about q, finally this will allow us to know the magnitude of
`q_(2)`

`q_(1)` exerts a force on q in +y direction, and
`q_(2)` exerts a force on q in -y direction.

`F_(1)=(kq_(1) q )/(d^2)\\F_(1)=((8.99*10^9)(25*10^(-6)C)(8.4*10^(-6)C))/((0.18m)^2)=58.26 N\\`

The net force on q is:

`F_(T)=F_(1) - F_(2)\\25N=58.26N-F_(2)\\F_(2)=58.26N-25N=33.26N\\\mid F_(2) \mid=(kq_(2)q)/(d^2)`

Rewriting for
`q_(2)`:

`q_(2)=(F_(2)d^2)/(kq)\\q_(2)=(33.26N(0.34m)^2)/(8.99*10^9(Nm^2)/(C^2)(8.4*10^(-6)C))=50.91*10^(-6)C=50.91 \mu C`