Question

Three charges are arranged as shown in the diagram.

The magnitude of the net electrical force acting on the +6 µC charge, rounded to the tenths place, is
N.

Answer 1

Answer: 5.7

Explanation: on edge its 5.7

Answer 2

Force due to -8 uC charge is given as

`F_1 = (kq_1q)/(r^2)`

`F_1 = ((9* 10^9)(8* 10^(-6))(6 * 10^(-6)))/(0.3^2)`

`F_1 = 4.8 N`

Force due to -5 uC charge is given as

`F_2 = (kq_2q)/(r^2)`

`F_2 = ((9* 10^9)(5* 10^(-6))(6 * 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 = √(F_1^2 + F_2^2)`

`F = √(3^2 + 4.8^2)`

`F = 5.66 N`

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