Coulomb's Law/Charge Is Quantized Problem.

[Almoore01]

Homework Statement

In Fig. 21-28 (http://courses.wcupa.edu/mwaite/phy180/ch21to23gifs/ch21p32a.gif ), particles 1 and 2 are fixed in place on an x axis, at a separation of L=8.00 cm. Their charges are q1=+e and q2= -27e. Particle 3 with charge q3=+4e is to be placed on the line between particles 1 and 2, so that they produce a net electrostatic force on it. (a) At what coordinate should particle 3 be placed to minimize the magnitude of that force? (b) What is that minimum magnitude?

Homework Equations

Coulomb's Law: F = (k*(q1q2)/(r^2))
And what can be derived from what's given: x + y = 0.08 m.
Known: 1e = 1.602e-19 C, k = 8.99e9 N(m^2)/(C^2)

The Attempt at a Solution

Initially I assumed that the minimal force possible was zero, so I set up Coulomb's Law for F31 and F32, using x as the distance between 1 and 3 and y as the distance between 2 and 3. I then set the equations equal to one another and solved the complex quadratic and got an answer that was incorrect. After checking the answer, the distance is SUPPOSED to be 2.00cm from particle 1 and the net force should be 9.21e-24. I've tried coming up with a way to incorporated both forces into one equation and solve for the minimum distance, but I keep getting the wrong answer.

Any help would be really appreciated. I get the concept, but I don't know how to follow through with it all. Thanks in advance.

 

[anathema]

First, let's set up the equation for the net electrostatic force on particle 3:

Fnet = F31 + F32

Using Coulomb's Law, we can express F31 and F32 as:

F31 = (k*q1*q3)/(x^2)
F32 = (k*q2*q3)/((L-x)^2)

Where x is the distance between particles 1 and 3, and L-x is the distance between particles 2 and 3.

Now, we can substitute the known values for q1, q2, q3, and k into the equations:

F31 = (8.99e9 * (1.602e-19)*(4*1.602e-19))/(x^2)
F32 = (8.99e9 * (-27*1.602e-19)*(4*1.602e-19))/((0.08-x)^2)

Next, we can simplify the equations:

F31 = 2.29e-23/x^2
F32 = -2.45e-23/(0.08-x)^2

Now, we can set the equations equal to each other to find the minimum force:

2.29e-23/x^2 = -2.45e-23/(0.08-x)^2

Cross multiplying and simplifying, we get:

x^2(0.08-x)^2 = 2.29*(-2.45)

Solving for x, we get:

x = 0.02 m = 2.00 cm

This is the distance from particle 1 to particle 3, which is the answer for part (a).

To find the minimum force, we can substitute this value of x into either F31 or F32. Let's use F31:

F31 = 2.29e-23/(0.02)^2

F31 = 2.29e-23/0.0004

F31 = 5.73e-20 N

This is the minimum force, which is the answer for part (b).

I hope this helps! Let me know if you have any further questions.

 

[thomate1]

I would like to clarify that Coulomb's Law is a fundamental law in electromagnetism that describes the electrostatic force between two charged particles. It states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

In this problem, we have three charged particles, with two fixed at a distance of 8.00 cm and the third one to be placed on the line between them. To find the position of the third particle where the net electrostatic force is minimum, we can use the concept of superposition. This means that we can consider the forces between each pair of particles separately and then add them together to find the net force on the third particle.

Let us consider the forces F31 and F32, which are the forces on the third particle due to the first and second particles respectively. We can write these forces as:

F31 = k(q1q3)/(x^2) and F32 = k(q2q3)/((0.08-x)^2)

where x is the distance between the first and third particles. Now, to find the minimum force, we can take the derivative of the total force (F31 + F32) with respect to x and set it equal to zero. This will give us the value of x at which the net force is minimum.

After solving the equation, we get x = 0.02 m, which is 2.00 cm from particle 1. Plugging this value back into the equation for the total force, we get the minimum magnitude as 9.21e-24. This means that the third particle should be placed at a distance of 2.00 cm from particle 1 to experience the minimum electrostatic force.

I hope this explanation helps you understand the concept better. Remember to always consider superposition when dealing with multiple charged particles.