Where Should Particle 3 Be Placed to Nullify Electrostatic Forces?

datenshinoai
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I'm completely stumped on three questions, please help...

1. Particle 1 of charge +1x10^-6C and particle 2 of charge -3.0x10^-6C are held at separation L=10.0cm on an x axis. If particle 3 of unknown charge q3 is to be located such that the net electrostatic force on it from particles 1 and 2 are sero, what must be the x and y coordinates of particle 3?

I know I need to use Coulomb's law, but since I don't know the unknown charge, how do I begin?

2. For particles are fixed in place and have charges q1=q2 = +5e, q3 = +3e, and q4 = -12e. They are all at a distance d = 5.0x10^-6m. What is the magnitude of the net electric field at point P due to the particles?
q4
|
|
q3
|
|
q1------P------q2​

I found that q1=q2 = 5e/[(4piEnot)(d)] so total = 2xans.

q3 = +3e/[(4piEnot)(d)]

q4 = -12e/[(4piEnot)(d)]

Do I just add it to get the net electric field or am I just not understanding the problem?

3. To charged particles -q = -3.0x10^-19C at x = -3.00m and q = 3.0x10^-19C at x = 3.00m. What are the magnitude and direction (relative to the positive direction of the x-axis) of the net electric field produced at point P at y = 4.00m?

I found E to be 2[3.2x10^-19/(4piEnot(3.0)^2)]sin 90, but it's not giving me the answer... does it have anything to do with the 4.00m? If so, how do I incorporate it in?

Thanks for taking the time to look at this!
 
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datenshinoai said:
I'm completely stumped on three questions, please help...

1. Particle 1 of charge +1x10^-6C and particle 2 of charge -3.0x10^-6C are held at separation L=10.0cm on an x axis. If particle 3 of unknown charge q3 is to be located such that the net electrostatic force on it from particles 1 and 2 are sero, what must be the x and y coordinates of particle 3?

I know I need to use Coulomb's law, but since I don't know the unknown charge, how do I begin?

2. For particles are fixed in place and have charges q1=q2 = +5e, q3 = +3e, and q4 = -12e. They are all at a distance d = 5.0x10^-6m. What is the magnitude of the net electric field at point P due to the particles?
q4
|
|
q3
|
|
q1------P------q2​

I found that q1=q2 = 5e/[(4piEnot)(d)] so total = 2xans.

q3 = +3e/[(4piEnot)(d)]

q4 = -12e/[(4piEnot)(d)]

Do I just add it to get the net electric field or am I just not understanding the problem?

3. To charged particles -q = -3.0x10^-19C at x = -3.00m and q = 3.0x10^-19C at x = 3.00m. What are the magnitude and direction (relative to the positive direction of the x-axis) of the net electric field produced at point P at y = 4.00m?

I found E to be 2[3.2x10^-19/(4piEnot(3.0)^2)]sin 90, but it's not giving me the answer... does it have anything to do with the 4.00m? If so, how do I incorporate it in?

Thanks for taking the time to look at this!

1) Just assume an arbitrary charge q. Do your Coulomb's law and you will find that the q drops out in the end. In order to do the vector addition though, you will need to know if the charge is positive or negative. This, too, should not matter in the end, so just assume that q is positive or negative. If you aren't convinced that it doesn't matter, do the problem both ways and see for yourself.

2) What does "q1=q2 = 5e/[(4piEnot)(d)]" mean? q1 and q2 are charges, right? I think that you are trying to write an expression for the electric field here... The electric field is E=q/(4\pi \epsilon_0 r^2). This is a vector, so you need to add them vectorally for each charge.

3) Again, this is a vector addition problem. Break E from each charge into componets and add the x componets together, etc.

-Dan
 
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