Question

Three charges are arranged as shown in the diagram.

Answer 1

Force due to -8 uC charge is given as

$F_1 = \frac{kq_1q}{r^2}$

$F_1 = \frac{(9\times 10^9)(8\times 10^{-6})(6 \times 10^{-6})}{0.3^2}$

$F_1 = 4.8 N$

Force due to -5 uC charge is given as

$F_2 = \frac{kq_2q}{r^2}$

$F_2 = \frac{(9\times 10^9)(5\times 10^{-6})(6 \times 10^{-6})}{0.3^2}$

$F_2 = 3 N$

Now we know that two forces on 6 uC charge is perpendicular to each other

so we will have

$F = \sqrt{F_1^2 + F_2^2}$

$F = \sqrt{3^2 + 4.8^2}$

$F = 5.66 N$

so net force due to both charges on 6 uC is 5.66 N

Answer 2

Answer: 5.7

Explanation: on edge its 5.7