Coulomb's Law - Calculating Net Force

In summary, Neonerd attempted to solve a problem involving charges of +70, +48, and -80 µC, but ended up with an incorrect answer. He was helped by another user, haruspex, who explained that the units of the constant in the equation, F, should be shown. Neonerd then solved the problem using the same equation but with the units of K, which is 9x10^9 Nm^2/C^2.
  • #1
Valentine
2
0
Hello all,

I am new to the website and I have joined because I just started a Physics class in College and I've never done any sort of Physics and after my first day I was thrown this problem and have been looking at it for a while now and finally gave in for some sort of help.

1. Homework Statement


Particles of charge +70, +48, and -80 μC are placed in a line. The center one is 0.35 m from each of the others. Calculate the net force on each charge due to the other two.
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2. The attempt at a solution

Before posting, I looked to see if any other posts were similar and I had came across one from 2005 and I've looked at his math:
F = (8.99*10^-9) (48*10^-6) (70*10^-6) / 0.35^2

which comes out to be: 2.46*10^-16

then F = (8.99*10^-9) (70*10^-6) (-80*10^-6) / 0.7^2

which comes out to be: -3.48*10^-16

And the answer I get it: 1.4*10^-16

I understand the user, neonerd, had said that he had made a mistake with the -9 on (8.99*10^-9) however I am not making the connection between

"F = (8.99*10^-9) (48*10^-6) (70*10^-6) / 0.35^2
= 2.46*10^-16"

I understand the formula is F = K q1 q2 / r^2

My professor has k=9x10^9 Nm^2/Coul^2 instead of using k=8.99x10^9 but says that meters and C cancel out so I know why K is different of neonerd's work

So the formula would start out as
F=(8.99x10^9) q1 q2 / r^2.
R is the distance so 0.35m^2
F=(8.99x10^9) q1 q2 / 0.35m^2
I am not understanding why the q1 and q2 are (70*10^-6) (-80*10^-6) and how neonerd got (2.46x10^2N)

Again I apologize if my misunderstanding is hair-pulling to anyone I just have never seen this and I would appreciate any help.
 
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  • #2
9 is just a rounded version of 8.99, and the minus in the exponent was a mistake.
Valentine said:
I am not understanding why the q1 and q2 are (70*10^-6) (-80*10^-6)
with Coulomb there. The charges are defined to have those values in the problem statement (µC, "micro-coulomb", means 10-6 Coulomb).
If you want to calculate the force between the 70µC and the -80µC charge: the distance between the charges is not 0.35m there.
 
  • #3
Valentine said:
says that meters and C cancel out
They only cancel out if all the distances plugged in are in metres and all the charges plugged in are in Coulombs.
Valentine said:
F=(8.99x10^9) q1 q2 / r^2
That's not strictly correct. Since F, q1, q2 and r are represented symbolically at this point, their units are unspecified in the equation. Therefore the units of the constant should be shown:
F=(9x109) q1 q2 / r2 Nm2/C2
When the values are plugged in:
F=(9x109) (70μC)*(48μC) / r2 (Nm2/C2)
=(9x109) 70*(48) / r2 (Nm2/C2) μC2
= (9x109) 70*(48) / r2 Nm2 (μC/C)2
= (9x109) 70*(48) / r2 Nm2 (10-6)2
= (9x10-3) 70*(48) / r2 Nm2
= (9x10-3) 70*(48) / (0.35m)2 Nm2
= (9x10-3) 70*(48) / (0.35)2 N
Valentine said:
q1 and q2 are (70*10^-6) (-80*10^-6) and how neonerd got (2.46x10^2N)
neonerd got that number for the case where the charges are 48 and 70 μC.
 
  • #4
with Coulomb there. The charges are defined to have those values in the problem statement (µC, "micro-coulomb", means 10-6 Coulomb).
If you want to calculate the force between the 70µC and the -80µC charge: the distance between the charges is not 0.35m there.
Micro! I feel very slow for not realizing it was micro That explains why it was 10^-6. Thank you! Also, I would have to use 0.7m^2 correct if it was from q1 to q3?
I am trying the math and my numbers are not adding up. I am usually great with math but -- well I'll just show my work.

That's not strictly correct. Since F, q1, q2 and r are represented symbolically at this point, their units are unspecified in the equation.

I see! I had just crossed out mine but I think practicing it the way you're showing would be better for the future because I am sure it will come up and try to haunt me. I'll be ready for it though!

F= K q1 q2 / r^2
=(9*10^9N*m^2/C^2) (70*10^-6C) (48x10^-6C) / 0.32^2 (0.35^2 Because it is from q1 to q2)
= I ended up canceling out the C^2 and m^2 even though I understand haruspex's work and ended up with:
(9x10^-3N) (70) (48) / 0.35^2 which is similar to haruspex's work but I left the N in the K area.
=Multiplying the top I got 30.24N/0.35^2
=Finally I get 246.857

I thought that it was my signs and double check and triple checked but I am not getting -1.4x10^2 or any numbers around 140.

________________

Okay I am seeing why. I am needing to come up with k q1 q3/r^2 THEN add them (I think)

________________

Yes! After doing the equation for q1 q3 I added up the two and got the answer! Thanks mfb and haruspex for your patience and time! I appreciate it now to do the other two parts of this problem! :D
 
Last edited:
  • #5


Hello and welcome to the website! I am glad you joined to seek help with your physics problem.

Coulomb's Law is a fundamental law in electromagnetism that describes the force between two charged particles. The formula, as you correctly stated, is F = K q1 q2 / r^2, where K is the Coulomb's constant, q1 and q2 are the charges of the particles, and r is the distance between them.

In this problem, you have three particles with different charges (+70 μC, +48 μC, and -80 μC) placed in a line. To calculate the net force on each charge, you need to consider the force exerted by the other two particles on each of them.

Let's consider the first particle with a charge of +70 μC. The force exerted on it by the second particle with a charge of +48 μC can be calculated using the formula F = K q1 q2 / r^2. Plugging in the values, we get F = (8.99x10^9) (70x10^-6) (48x10^-6) / (0.35)^2 = 2.46x10^-16 N. This is the same value that neonerd got in their solution.

Similarly, the force exerted on the first particle by the third particle with a charge of -80 μC can be calculated using the same formula. Plugging in the values, we get F = (8.99x10^9) (70x10^-6) (-80x10^-6) / (0.35)^2 = -3.48x10^-16 N.

To find the net force on the first particle, we need to add these two forces together, taking into account their direction. Since the first force is positive and the second force is negative, the net force will be 2.46x10^-16 N - 3.48x10^-16 N = -1.02x10^-16 N. This is the same value that neonerd got in their final answer.

I hope this explanation helps you understand the problem and how to calculate the net force on each charge using Coulomb's Law. Keep practicing and don't hesitate to ask for help if you have any further questions. Good luck with your physics class!
 

Related to Coulomb's Law - Calculating Net Force

What is Coulomb's Law?

Coulomb's Law is a fundamental law in physics that describes the force between two electrically charged particles. It states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

How is net force calculated using Coulomb's Law?

To calculate the net force between two charged particles, we use the formula F = (k * q1 * q2) / d2, where F is the net force, k is the Coulomb's constant, q1 and q2 are the charges of the particles, and d is the distance between them.

What is the unit of measurement for force in Coulomb's Law?

The unit of measurement for force in Coulomb's Law is Newtons (N). This is because force is a vector quantity and is measured in terms of mass and acceleration, which is represented by the unit N.

Can Coulomb's Law be applied to all types of charged particles?

Yes, Coulomb's Law can be applied to all types of charged particles, including both positive and negative charges. It is a universal law that governs the force between any two charged particles.

What is the significance of Coulomb's constant in the formula for net force?

Coulomb's constant, denoted by k, is a constant that helps us convert the units of charges and distances into the correct units for force. It also helps us compare the strength of the forces between different charged particles.

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