Question

A negative charge, q1, of 6 µC is 0.002 m north of a positive charge, q2, of 3 µC. What is the magnitude and direction of the electrical force, Fe, applied by q1 on q2?

Answer 1

Answer:

Force,F = $4.05\times 10^4\ N$

Explanation:

Given that,

Charge 1, $q_1=-6\ \mu C=-6\times 10^{-6}\ C$

Charge 2, $q_2=+3\ \mu C=3\times 10^{-6}\ C$

Negative charge is 0.002 m north of a positive charge, distance between charges, d = 0.002 m.

Let $F_e$ is the magnitude of electrical force applied by q₁ on q₂. We know that the electrical force is given by the formula as follows :

$F_e=k\dfrac{q_1q_2}{d^2}\\\\F_e=9\times 10^9\times \dfrac{6\times 10^{-6}\times 3\times 10^{-6}}{(0.002)^2}\\\\F_e=4.05\times 10^4\ N$

So, the electrical force applied by q₁ on q₂ is $4.05\times 10^4\ N$ towards north.

Answer 2

Answer:

D) magnitude: 4 × 104 N

direction: north

Explanation:

on edg