Question

A particle with charge 8 µC is located on the

Answer 1

Answer: 34.65 N towards charge 8 μC.

Explanation:

The electrostatic force between two charges is given by:

$F =k \frac{q_1q_2}{r^2}$

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:

$F = 8.9875 \times 10^9 Nm^2/C^2 \times \frac{8\times 10^{-6}C\times (-7\times10^{-6}C)}{(0.12m)^2} = -34.95 N$

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:

$F' = 8.9875\times 10^9Nm^2/C^2\times \frac{(-3 \times 10^{-6}C)\times (-7\times 10^{-6}C)}{(0.8m)^2} = 0.295 N$

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.