Electric Field/Charges Physics Problem

[PCMarine]

I have a worksheet to do for my physics class, and the rest of the problems depend upon the first one which says:

Two point charges, Q1 and Q2, are located a distance 0.20 meters apart, as shown above. Charge Q1= +8.0 microCoulombs. The net electric field is zero at point P, located 0.40 meters from Q1 and 0.20 meters from Q2.

Determine the magnitude and sign of charge Q2.

I tried using Coulomb's law (Electric Field = k ((Q1*Q2)/r^2) to find Q2, but you don't know the Electric Field (aside from it being zero at point P).

Any help would be most appreciated, thanks.

 

[91TTZ]

You need to change the flux capacitor.

 

[PCMarine]

bump

 

[mindmaniac]

You need to set the two charges equal to each other. So (Q1/r1^2)=(Q2/r2^2). You can neglect coulombs constant since they will cancel each other out. And once you draw the diagram of the point charges you will see that Q2 has to be negative. Q2 should be (-)2 microcoulombs.

Q1- 2.0 m to left from Q2 - and P is 2.0 m to the right of Q2

Hope this helps

 

[PCMarine]

Thanks, that helps quite a bit!

 

[audi]

electrostatics is the hardest thing i've done in ap physics this year. that problem was rather simple though.

 

[chiwawa626]

Modern physics is so much easyer lol 🙂

 

[Random Variable]

Originally posted by: audi
electrostatics is the hardest thing i've done in ap physics this year. that problem was rather simple though.

Take a class in quantum statistical mechanics. It's starts out making at least some sense, but quickly devolves into such abstract concepts that your brain explodes.

 

[esun]

Q1 |<----d---->| Q2

d = 0.20 m
Q1 = +8.0 uC

Now you have to figure out where P is in relation to Q1 and Q2. Since it is 0.40 m from Q1 and 0.20 m from Q2, you know it must be 0.20 m to the right of Q2. You also know that since E is a vector, the component at P given by Q1 and Q2 must be pointing in the opposite directions and of equal magnitude.

So, getting them in opposite directions is easy enough. Q2 < 0. Done with that part. Now, to get the magnitude you simply add the contributions from Q1 and Q2. So you have:

k(Q1/0.4^2) + k(-Q2/0.2^2) = 0

This is also

Q1/0.4^2 = Q2/0.2^2

You have Q1, so plug in a solve. Q2 = 1/4*Q1 (which, conceptually, makes sense since Q2 is twice as close to P as Q1, and E is proportional to 1/r^2).


EDIT: Oh, and you got Coulomb's Law wrong. Electric field = k*Q/r^2. Coulomb's Law states that Electric FORCE = k*Q1*Q2/r^2.