Question

Two negative charges, Q = -50 μC are placed on the x-axis at x = +0.5 m and x = -0.5 m.

(Questions below.)

Answer 1

a) How much energy does it take to assemble these two charges? Express your answer using three significant figures and in the unit J.

b) You place an unknown charge on the y-axis at y = 1 m and find that the electric potential is 0 V at the origin. What is the unknown charge? Answer in μC.

c) If you place a -1 μC charge at the origin, what is the x-component of the force on it from the other three charges? Answer in x-hat N. <<< Found this to be 0 >>>

d) If you place a -1 μC charge at the origin, what is the y-component of the force on it from the other three charges? Answer in y-hat N.

Answer 2

For part a) I used the definition of work in relation to potnetial: W(or energy)=qV.
For part d) it is very simple it will be a repelling force equal to F=\[kq_1q_2/r^2 \] where \[q_1=-1 \mu C\] and [q_2=-450 \mu C\] and k is equal to coulomb's constant and r is equal to 1.00 meter. This Force should be in the negative y-hat direction.

Answer 3

Wow, you really helped me out to visualize the question better! I should really draw these things out more...!
For a), I did end up using the right equation, but I didn't square it for some reason. But that's what I did wrong.

For b), it's telling me it's wrong, though.
If it helps, my friend had the same problem but with different numbers: -25μC and 0.5m. She couldn't figure it out and the computer gave the solution of 100μC, so I'm sure you are on the right track.

Answer 4

I'm trying to work out d.
It's telling me I'm wrong too.

This is how I'm setting it up:

F=(k)(q1)(q2)/(r^2)
F=(9*10^9)(-1*10^-6)(-450*10^-6)/(1.00^2)
F= 4.05 μC

Again, my friend had the same units as above (-25μC and 0.5m) and her answer was 2.70μC... So I'm guessing it's similar as well?

Answer 5

The answer to b) was 200μC and c) was 1.80.

Thanks for the help with some of it -- I'll give you a medal anyway.