shaqadry
Two point charges of equal magnitude are
7.5 cm apart. At the midpoint of the line
connecting them, their combined electric field
has magnitude of 45 N/C. Find the magnitude
of the charges.
Pls show me the complete calculation
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Mashy
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first tell me what is the formula to calculate electric field due to a point charge?
shaqadry
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E=F/q=kq/r^2
Mashy
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ok.. good.. tell me.. do you think the charges should be of same nature or opposite nature?? or both possible? doesn't matter?
shaqadry
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i think it has the same nature since the magnitude of electric field is high
Mashy
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ok.. if the charges are of same nature.. then think the direction of the electric field due to each charge at the mid point? will they be same direction or opposite direction?
Mashy
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|dw:1354890053415:dw|
shaqadry
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i think same direction
shaqadry
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i dont know what happens at the center :-/ the radius is halved i guess?
Mashy
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how is that?? the direction of electric field due to a positive charge is always AWAY from the charge!!
Mashy
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|dw:1354890336186:dw|
shaqadry
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okay
Mashy
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|dw:1354890474542:dw|
Mashy
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now.. Ea + Eb = the required answer.. !!... Ea you can find.. Eb you can find.. add them equate to required answer.. get Q
shaqadry
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the net force is 0?
shaqadry
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i did try that but i cant seem to get the correct anwers. the answer is 3.5 x 10^-12
Mashy
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No now.. the net force will add up and = 45 N/C
Mashy
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net field*
Mashy
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\[2Q/4\pi \epsilon r ^{2}= 45\]
Mashy
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r is .075/ 2 cause the mid point is half the distnace
shaqadry
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i got the answer! thank you very much :)
Mashy
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your welcome!.. i could have given the answer before.. but i wanted you to think and get it :P
shaqadry
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i know it helps a lot! so thank you!
Mashy
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anytime :)!