Electrostatic Force: +3.2 C & -1.6 C Magnitude of Force

In summary, the question is asking which statement correctly describes the magnitude of the electric force acting on two charges, +3.2 C and -1.6 C, separated by a distance of radius r. The answer is c), as the force on q1 and q2 would be equal regardless of which charge is considered first. This is due to Newton's third law, which states that for every action there is an equal and opposite reaction.
  • #1
Luke0034
26
1

Homework Statement



I have a question that I think I know, but it is kind of confusing me a little bit. The problem is as following:

Consider a charge of +3.2 C and a charge of -1.6 C separated by a distance of radius r. Which of the following statements correctly describes the magnitude of the electric force acting on the two charges?

a) The force on q1 has a magnitude that is twice that of the force on q2.
b) The force on q2 has a magnitude that is twice that of the force on q1.
c) The force on q1 has the same magnitude as that of the force on q2.
d) The force on q2 has a magnitude that is four times that of the force on q1.
f) The force on q1 has a magnitude that is four times that of the force on q2.

Homework Equations



F = kq1q2/r^2

The Attempt at a Solution



I think the answer is c, because no matter if you plug on kq1q2/r^2 or kq2q1/r^2, you'll get the same amount of force. Am I doing this correctly. I'm sure if I'm thinking of this problem in the right way. I know that q1 would emit a stronger electric weak, and q2 would have a weaker electric field, but I'm not sure how that relates to the amount of force q1 applies to q2, or q2 applies to q1. Is there a way to discriminate the force that q1 applies to q2, and q2 on q1?? Or are they the same?
 
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  • #2
Luke0034 said:

The Attempt at a Solution



I think the answer is c, because no matter if you plug on kq1q2/r^2 or kq2q1/r^2, you'll get the same amount of force. Am I doing this correctly. I'm sure if I'm thinking of this problem in the right way. I know that q1 would emit a stronger electric weak, and q2 would have a weaker electric field, but I'm not sure how that relates to the amount of force q1 applies to q2, or q2 applies to q1. Is there a way to discriminate the force that q1 applies to q2, and q2 on q1?? Or are they the same?

It is c). There's also a thing called Newton's third law to guide you!
 
  • #3
PeroK said:
It is c). There's also a thing called Newton's third law to guide you!

Thank you, that's makes a lot of sense. After reading a little bit, I understand Newton's third law in an intuitive sense now.
 

Related to Electrostatic Force: +3.2 C & -1.6 C Magnitude of Force

1. What is electrostatic force?

Electrostatic force is the force that exists between two electrically charged particles or objects. This force can be either attractive or repulsive, depending on the type of charges present.

2. How is electrostatic force calculated?

The magnitude of electrostatic force is calculated using Coulomb's Law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

3. What is the unit of measurement for electrostatic force?

The SI unit for electrostatic force is the Newton (N). However, for smaller charges, the unit of measurement is often expressed in Coulombs (C).

4. How does the magnitude of force change with distance?

According to Coulomb's Law, as the distance between two charged particles increases, the force between them decreases. This is because the inverse square relationship between distance and force means that the force becomes weaker as the distance increases.

5. What is the effect of increasing or decreasing the magnitude of charges on the electrostatic force?

Increasing the magnitude of charges on two particles will result in a stronger electrostatic force between them. Conversely, decreasing the magnitude of charges will result in a weaker force. This is because the force is directly proportional to the charges involved.

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