Question

What is the force and direction in q1 due to the other two charges (q2 and q3)?

a) 10.0 UC, attractive
b) 10.0 UC, repulsive
c) 2.0 UC, attractive
d) 2.0 UC, repulsive

Answer 1

The force on
`\(q_1\)` due to the other two charges
`(\(q_2\) and \(q_3\))` is 2.0 micro-Coulombs
`(\(\mu C\))` and attractive.

Step-by-step explanation:

To calculate the force between charges, we use Coulomb's Law, which states that the force (F) between two charges is given by:

`\[ F = (k \cdot |q_1 \cdot q_2|)/(r^2) \]`

where k is Coulomb's constant,
`\(q_1\)` and
`\(q_2\)` are the magnitudes of the charges, and r is the separation between the charges.

In this case, \
`(q_1 = 4 \, \mu C\), \(q_2 = -3 \, \mu C\), and \(q_3 = 5 \, \mu C\).` The force between \
`(q_1\)` and
`\(q_2\)` is attractive:

`\[ F_(q_1q_2) = (k \cdot |q_1 \cdot q_2|)/(r_(12)^2) \]`

The force between
`\(q_1\) and \(q_3\)` is repulsive:

`\[ F_(q_1q_3) = (k \cdot |q_1 \cdot q_3|)/(r_(13)^2) \]`

The net force on
`\(q_1\)` due to
`\(q_2\)` and
`\(q_3\)` is the vector sum of these individual forces. Since the magnitudes of the forces are equal, the net force is attractive and has a magnitude of
`\(2.0 \, \mu C\).`