Question

A charge of +3.00 μC is located at the origin, and a second charge of -2.00 μC is located on the x-y plane at the point (50.0 cm, 60.0 cm). 1) Determine the x-component of the electric force exerted by the -2.00 μC charge on the +3.00 μC charge. (Express your answer to three significant figures.)

Answer 1

Answer:

0.0567 N

Explanation:

q1 = 3 micro coulomb

q2 = - 2 micro coulomb

OB = 50 cm

AB = 60 cm

By using Pythagoras theorem in triangle OAB

$OA^{2}=OB^{2}+AB^{2}$

$OA^{2}=50^{2}+60^{2}=6100$

OA = 78.1 cm

By using the Coulomb's law, the force on 3 micro coulomb due to -2 micro coulomb is

$F=\frac {Kq_{1}q_{2}}{OA^{2}}= \frac{9 \times 10^{9}}\times 3\times 10^{-6} \times 2 \times 10^{-6}{0.781^{2}}$

F = 0.0885 N

The horizontal component of force is

= F CosФ = $F\times \frac{OB}{OA}$

= 0.0885 x 50 / 78.1 = 0.0567 N