How Does Charge Polarity Affect Electrostatic Equilibrium Position?

In summary, the problem involves finding the position where a third point charge, q, can be placed on the real axis so that the net electrostatic force on it is zero. The problem is solved using the equation k(q1)(q2)/d^2=k(q2)(q3)/(D-d)^2, where d is the distance from q1 to q, D is the distance from q to q2, and k is a constant. By rearranging the equation and solving for d, we can determine that the charge must be placed in the region between Q1 and Q2, specifically at a distance of 1.643 m on the real axis. In the second part of the problem, the sign of Q
  • #1
badd99
21
0

Homework Statement



An electric point charge of Q1 = 5.23 nC is placed at the origin of the real axis. Another point charge of Q2 = 1.89 nC is placed at a position of p = 2.63 m on the real axis. At which position can a third point charge of q = -2.62 nC be placed so that the net electrostatic force on it is zero?

1.643 m (SOLVED)


Let the sign of Q2 be changed from positive to negative. At which position can the point charge q be placed now so that the net electrostatic force on it is zero?


The Attempt at a Solution



Okay so I set up k(q1)(q2)/d^2=k(q2)(q3)/(D-d)^2 where D = 2.63m

I then used the quadratic and since there is no b, I used 0.

I got:
7.12E-6d^2+0d-3.62E-5 and did the quadratic to get 2.25m but this is not right?

Help? Thanks!
 
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  • #2
The first thing to figure out is which region the charge must be placed: to the left of Q1, between Q1 and Q2, or to the right of Q2. Only one region will give you a sensible answer.
 
  • #3
so the answer should be negative then? I am still having trouble ugh.
 
  • #4
badd99 said:

The Attempt at a Solution



Okay so I set up k(q1)(q2)/d^2=k(q2)(q3)/(D-d)^2 where D = 2.63m
Define d and D more clearly.

Better still, pick a region and define your distances so they are all positive.
 
  • #5


I would say that your attempt at a solution is on the right track, but there may be a mistake in your calculation. It is important to double check your equations and make sure all units are consistent. Also, when using the quadratic formula, make sure to use the correct sign for the discriminant (in this case, it should be positive since the force should be zero). Another helpful tip is to draw a diagram and label all the variables to help visualize the problem. Keep in mind that the position of the third point charge can be on either side of the other two charges, so make sure to consider both cases when solving for the position. It may also be helpful to consult with a classmate or teacher for additional clarification. Keep up the good work!
 

Related to How Does Charge Polarity Affect Electrostatic Equilibrium Position?

1. What is electrostatic force position?

Electrostatic force position refers to the position of a charged object or particle in relation to another charged object or particle, which affects the strength and direction of the electrostatic force between them.

2. How is electrostatic force position calculated?

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

3. What factors affect electrostatic force position?

The factors that affect electrostatic force position include the magnitude and sign of the charges, as well as the distance between them.

4. How does electrostatic force position affect the movement of particles?

Electrostatic force position affects the movement of particles by determining the direction and strength of the electrostatic force between them. Oppositely charged particles will be attracted to each other, while particles with the same charge will repel each other.

5. What is the significance of electrostatic force position in everyday life?

Electrostatic force position plays a crucial role in many everyday phenomena, such as static electricity, lightning, and the behavior of charged particles in electronic devices. It also helps explain the behavior of atoms and molecules in chemical reactions.

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