Answer to Question #299325 in Electric Circuits for ema

Question

A test charge of positive 2 micro-Coulomb is placed halfway between a charge of positive 6 microCoulomb and another positive 4 micro-Coulomb separated by 10 centimeter.

a. What is the magnitude of the force on the test charge?

b. What is the direction of this force (away from or toward the positive 6 micro-Coulomb 

Answer 1

Solution;

We know;

$F=\frac{kq_1q_2}{r^2}$

a)

Force on the test charge due to the 6uC;

$F_6=\frac{8.99×10^9×2×10^{-6}×6^{-6}}{0.05^2}=43.152N$

Force on the test charge due to the 4uC;

$F_4=\frac{8.99×10^9×2×10^{-6}×4×10^{-6}}{0.05^2}=28.768N$

The net charge is;

$F_{net}=43.152-28.768=14.384N$

(b)

Since the 6uC exerts the greater force,net force on the test charge is toward the 6uC.