Answer:
0.875 nC, positive
Explanation:
The electrostatic force between two charges is given by:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where k is the Coulomb's constant, q1 and q2 are the two charges, r the separation between them
The force is attractive if the two charges have opposite signs and repulsive if the two charges have same sign
Here we have:
- Charge q3 (positive) at position x = 0
- Charge q1 at position x = 2.50 cm
- Charge q2 (negative) at position x = 5.00 cm
The force exerted by q2 on q3 is attractive, so q3 is pulled towards the positive x direction - therefore in order to have a net force of zero on q3, the force exerted by q1 on q3 must be repulsive, which means that q1 must be positive.
Concerning the magnitude of the charge, we can find it by requiring that the vector sum of the two forces is zero. Therefore:
[tex]F_{13} + F_{23} = 0[/tex]
Since the forces are in line, this becomes
[tex]k\frac{q_1 q_3}{r_{13}^2}+k\frac{q_2 q_3}{r_{23}^2}=0[/tex]
And by solving for q1, we find:
[tex]q_1=-\frac{q_2 r_{13}^2}{r_{23}^2}=-\frac{(-3.5\cdot 10^{-9}C)(0.025m)^2}{(0.05 m)^2}=8.75\cdot 10^{-10} C=0.875 nC[/tex]
