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

Answer:

$50.91 \mu C$

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.

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

The net force on q is:

$F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}$

Rewriting for $q_{2}$:

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