97.8k views
1 vote
A 7.45 nC charge is located 1.66 m from a 4.22 nC point charge. (a) Find the magnitude of the electrostatic force that one charge exerts on the other.(b) Is the force attractive or repulsive?i. attractiveii. repulsive

1 Answer

2 votes

Answer:

a.
F=1.03* 10^(-7)N

b.Repulsive

Step-by-step explanation:

We are given that


q_1=7.45nC=7.45* 10^(-9)C


1nC=10^(-9)C


q_2=4.22nC=4.22* 10^(-9)C

r=1.66m

We know that

Electrostatic force =
F=k(q_1q_2)/(r^2)

r=Distance between
q_1\;and\;q_2

k=Constant=
9* 10^9Nm^2C^(-2)

Using the formula

a.The magnitude of the electrostatic force=
F=(9* 10^9* 7.45* 10^(-9)* 4.22* 10^(-9))/((1.66)^2)

The magnitude of the electrostatic force=
F=1.03* 10^(-7)N

b.Both charge are positive .We know that when like charges repel each other and unlike charges attract to each other.

Therefore, the force between given charges is repulsive.

User NameOfTheRose
by
5.1k points