Answer:
(a) Magnitude: 14.4 N
(b) Away from the +6 µC charge
Explanation:
As the test charge has the same sign, the force that the other charges exert on it will be a repulsive force. The magnitude of each of the forces will be:
[tex]F_e = K\frac{qq_{test}}{r^2}[/tex]
K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q and qtest is the charge of the particles, and r is the distance between the particles.
Let's say that a force that goes toward the +6 µC charge is positive, then:
[tex]F_e_1 = K\frac{q_1q_{test}}{r^2}=-9*10^9 \frac{Nm^2}{C^2} \frac{6*10^{-6}C*4*10^{-6}C}{(0.1m)^2} =-21.6 N[/tex]
[tex]F_e_2 = K\frac{q_2q_{test}}{r^2}=9*10^9 \frac{Nm^2}{C^2} \frac{2*10^{-6}C*4*10^{-6}C}{(0.1m)^2} =7.2 N[/tex]
The magnitude will be:
[tex]F_e = -21.6 + 7.2 = -14.4 N[/tex], away from the +6 µC charge
