Respuesta :
Answer: 34.65 N towards charge 8 μC.
Explanation:
The electrostatic force between two charges is given by:
[tex] F =k \frac{q_1q_2}{r^2}[/tex]
where, k is the Coulomb constant = 8.9875 × 10⁹ N.m²/C²
q₁ and q₂ are the two charges separated by distance r.
The distance between charges 8 μC and -7 μC is r = 2 cm -(-10 cm) = 12 cm = 0.12 m
The force between these charges is:
[tex]F = 8.9875 \times 10^9 Nm^2/C^2 \times \frac{8\times 10^{-6}C\times (-7\times10^{-6}C)}{(0.12m)^2} = -34.95 N[/tex]
Negative sign implies it is an attractive force.
The distance between -3 μC charge and -7 μC charge is r' = 10 cm -2 cm = 8 cm = 0.8 m.
The electrostatic force between these charges is:
[tex]F' = 8.9875\times 10^9Nm^2/C^2\times \frac{(-3 \times 10^{-6}C)\times (-7\times 10^{-6}C)}{(0.8m)^2} = 0.295 N[/tex]
It is a repulsive force.
Net force on the -7 μC charge is:
Fn = F + F'
we can add them directly as they are acting in one direction along the x-axis.
Fn = -34.95 N + 0.295 = -34.65 N
Thus, the net force is attractive in nature and it is towards charge 8 μC.
