Aparticle with charge 8 µc is located on the
x-axis at the point −10 cm , and a second
particle with charge −3 µc is placed on the
x-axis at 10 cm .
what is the magnitude of the total electrostatic force on a third particle with charge
−7 µc placed on the x-axis at 2 cm ? the
coulomb constant is 8.9875 × 109 n · m2
/c
2
.
answer in units of n.

34.65 N towards charge 8 μC.

Explanation:

The electrostatic force between two charges is given by:

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:

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:

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.

unbalanced force

explanation: