Question

Three charges, each of magnitude 2 nC, are at separate corners of a square of edge length 2 cm. The two charges at opposite corners are positive, and the other charge is negative. Find the force exerted by these charges on a fourth charge q +3 nC at the remaining (upper right) corner. (Assume the +x axis is directed to the right and the ty axis is directed upward.)

Answer 1

Answer:

$F_{net} = 1.23 \times 10^{-4} N$

Explanation:

Force due to two positive charges on 3 nC is given as

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

$F_1 = F_2 = \frac{(9\times 10^9)(2nC)(3 nC)}{0.02^2}$

$F_1 = F_2 = 1.35 \times 10^{-4} N$

now the force on 3 nC charge due to opposite charge - 2 nC is given as

$F_3 = \frac{(9\times 10^9)(2nC)(3 nC)}{2(0.02)^2}$

$F_3 = 0.675 \times 10^{-4} N$

now the force F3 is an attractive force while F1 and F2 is repulsive nature force

so we will have

$F_{net} = \sqrt2 F_1 - F_3$

$F_{net} = \sqrt2 (1.35 \times 10^{-4}) - (0.675 \times 10^{-4})$

$F_{net} = 1.23 \times 10^{-4} N$