Two particles having charges q1 = 0.600 nC and q2 = 7.50 nC are separated by a distance of 1.60 m . At what point along the line connecting the two charges is the total electric field due to the two charges equal to zero?
The answer is supposed to be the charge in meters from q1. I'm down to my last try on this one as well and I always felt like I was doing it right
The way to solve is to write two equations for the electric field strength. One for q1 and the other for q2. The distances from each charge must always equal 1.6 m. So one of the distances can be expressed as the other. Then the two equations can be combined into one and solved for one of the distances. I think the field strength is inversely proportional to the square of the distance away from the charge.
I got .328 m but it was wrong
I think the zero field point is definitely closer to the q1 charge. Am I right about the field being proportional to inverse square of distance away?
I meant to say earlier that the sum of the two distances must equal 1.6 m. Could that be a factor in your answer?
I'm not really sure. I've tried 8 other solutions and got all the answers wrong and don't even know what I'm doing anymore
Sorry you missed the answer. I have a freshman physics book. I can review the relations between distance and field strength. Maybe then I could do a calculation for you. Give me about 20 min.
Here is what I get: F = q/4pier^2 The 4pie can be canceled out because it is on both sides of the equation. The setup equation is 0.6/(1.6-r)^2=7.5/r^2 r is the distance from the 7.5 nC charge. Solving this, r = 1.247 and the other distance from the weaker charge is 0.353 m. I hope this helps, maybe for a later problem you might have.
So, the answer would be .353 m from the charge q1?
Yes
Thanks! You're a lifesaver! I got one more question if you could? I really don't know what the question is asking.
Okay. Give me the question.
I'll open a new one
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