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October 24, 2014

October 24, 2014

Posted by **jennifer** on Tuesday, February 2, 2010 at 2:42pm.

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 3:16pm1. 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.

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