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1. A + 2.2 * 10 ^ -9charge is on the x- axis @ x = 1.5 m, a + 5.4 * 10 ^ -9 C charge is on the x- axis @ x= 2.0 m, and a +3.5 * 10 ^ -9 C charge is @ the origin. Find the net force on the charge at the origin?
2. A charge q1 of -6.00 * 10 ^ -9 and a charge q2 of -3.00 * 10 ^ -9 C are seperated by a distance of 60 cm. Where could a third charge be places so that the net elctric force on it is zero ?
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I'm in no mood to crunch the numbers for you. But it's easy enough. It's just a matter of using Coulomb's law, F = k* q1 * q2 / r^2.
1) Use the magnitude of the first charge and the charge at the origin to find their attractive force. Then use the magnitude of the second charge and the charge at the origin to find their attractive force. Then add these numbers together. Make sure your direction ends up correct; since the charges are repelling, the force on the charge at the origin should be negative.
2) A little more tricky. Let's just set up a coordinate system. Any coordinate system will do. Let's say that q1 is at x = 0 and q2 is at x = 0.60. We could make up an imaginary charge for the third particle (or just leave it as a variable and find that it will cancel), but it's easier if you realize that the force on it will be zero only if its location has no electric field. So we're looking for the spot here where E = 0. Remembering the electric field formula, we have E = k * q / r^2. We can then set up the following equation:
E1 + E2 = 0
(k * q1 / r1^2) + (k * q2 / r2^2) = 0
(k * q1 / x^2) + (k * q2 / (0.60 - x)^2) = 0
Then it's just a matter of plugging in our numbers for k and the charges and solving for x, which will locate the third charge on the coordinate system we made up. Since this coordinate system wasn't given in the problem, make sure your answer explains the location and does not just give the number you find for x.