Three equal charges are placed in such a way that they form a right triangle with sides 3 cm , 4 cm , 5 cm . Each charge has a magnitude of 2 × 1 0 − 6 C .
Find the net force acting on the charge at the right angle.
Details and assumptions:
Take 4 π ϵ 0 1 = 9 × 1 0 9 Nm 2 / C 2 .
Round your answer to the nearest integer.
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Would someone like to elaborate the solution a little bit more?
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Which part exactly confuses you?
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Sravanth, I think I should've uploaded a diagram of this system. You can add that picture in the problem statement, but that will make the problem easy.
Lemme see what I can do.
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@Muhammad Arifur Rahman – Yeah, that's why I didn't put a diagram.
I will be clarifying it soon, OK!
@Sravanth - Please modify the unit of electrostatic constant in the problem statement. It would be N m 2 / C 2 instead of N C 2 / m 2 !
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Thnaks! I've edited it!
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So nice! But you still haven't supported my solution with your vote.
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@Muhammad Arifur Rahman – Oh my! I forgot that, sorry!
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@Sravanth C. – ……Waiting for your next awesome problem!
Oh . I forgot to convert the measurments
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At the corner, two forces are acting making an angle of 90 degrees between them. And the other two charges are 3cm and 4cm away from the corner charge.
Now given, k = 4 π ϵ 0 1 = 9 × 1 0 9 N m 2 / C 2 q = 2 μ C r 1 = 3 c m = 0 . 0 3 m r 2 = 4 c m = 0 . 0 4 m
So, F = ( r 1 2 k q 2 ) 2 + ( r 2 2 k q 2 ) 2 = k q 2 r 1 4 1 + r 2 4 1
Plugging the values, F = 4 5 . 8 9 3 N After rounding up to the nearest integer, F = 4 6 N