I have these problems that I keep doing and keep missing. If someone could please help me with what I am doing wrong!
Two charges, -25 µC and +15 µC, are fixed in place and separated by 3.4 m.
(a) At what spot along a line through the charges is the net electric field zero? Locate this spot relative to the positive charge. (Hint: The spot does not necessarily lie between the two charges.)
(b) What would be the force on a charge of +16 µC placed at this spot?
I did this a bunch of ways. One was saying E=kq/r.
So I got E=k(-25x10^-6)/x^2 + k(15x10^-6)/(3.4-x)^2 I had no idea where to go from there
Another way was (25)(x)^2 = 15(x-3.4)^2 From here I got the crazy answer of 4.164132563.
2. Two charges are placed between the plates of a parallel plate capacitor. One charge is +q1 and the other is q2 = +2.50 µC. The charge per unit area on each plate has a magnitude of σ = 2.00 10-4 C/m2. The force on q1 due to q2 equals the force on q1 due to the electric field of the parallel plate capacitor. What is the distance r between the two charges?
I used E=sigma/Eo which was 2x10^-4/8.85x10^-12. This came to 2.2598870.06
Then I did F=qE and F=kq1q2/r^2.
I set this = to each other with the q’s cancelling out and got 100. Which is wrong.
physics - bobpursley, Tuesday, February 2, 2010 at 3:16pm
1. one charge negative, one charge positive. You wont get a solution in between the charges, because the E is in the SAME direction, they are not in opposite direction, so they will never cancel.
So, the position of x is outside the charges. Let x be the distance from one, 3.4+x be the other distance.