Question

A positive charge, q1, of 5 µC is 3 × 10–2 m west of a positive charge, q2, of 2 µC. What is the magnitude and direction of the electrical force, Fe, applied by q1 on q2? magnitude: 3 N direction: east magnitude: 3 N direction: west magnitude: 100 N direction: east magnitude: 100 N direction: west

Answer 1

Answer:

magnitude: 100 N direction: east

Explanation:

The electrostatic force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the charges

In this problem, we have

$q_1 = 5 \mu C = 5\cdot 10^{-6} C$

$q_2 = 2 \mu C = 2\cdot 10^{-6} C$

$r=3\cdot 10^{-2} m$

Substituting into the equation, we find the magnitude of the force

$F=(9\cdot 10^9 N m^2 C^{-2})\frac{(5\cdot 10^{-6}C)(2\cdot 10^{-6}C)}{(3\cdot 10^{-2}m)^2}=100 N$

Concerning the direction, let's notice that:

- Both charges are positive, so the force between them is repulsive

- The charge q1 is west of the charge q2

- So, the force applied by q1 on q2 must be to the east (away from charge q1)

Answer 2

Answer:

Magnitude 100 N, direction East

Explanation:

Edge test answer confirmed