Answer :
The resultant force on the 8.00 nC charge is -0.010 N.
Using Coulomb's law, the electric force between two point charges is given by,
F = k * (q1 * q2) / r^2
where F is the force, k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between the charges.
The force on the 8.00 nC charge due to the -2.50 nC charge:
F1 = k * ((8.00 nC) * (-2.50 nC)) / (0.020 m)^2
= -0.090 N
Note that the negative sign indicates an attractive force, as the charges have opposite signs.
The force on the 8.00 nC charge due to the -4.00 nC charge,
F2 = k * ((8.00 nC) * (-4.00 nC)) / (0.030 m)^2
= -0.080 N
Add these forces vectorially. Since the forces are acting in opposite directions,
Fnet = F1 - F2
= -0.090 N - (-0.080 N)
= -0.010 N
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