Asked by Sarah
Identical +7.74 μC charges are fixed to adjacent corners of a square. What charge (magnitude and algebraic sign) should be fixed to one of the empty corners, so that the total potential at the remaining empty corner is 0 V?
I thought that this is what i should do. But I cant get the correct answer...
the total potential is
V = k[7.74μC/r + 7.74μC/√2 *r + q/r]
0 = k/a[(7.74μC+q)√2 + 7.74μC]
or (7.74μC+q)√2 = - 7.74μC
q = -7.74μC[1+1/√2]
I thought that this is what i should do. But I cant get the correct answer...
the total potential is
V = k[7.74μC/r + 7.74μC/√2 *r + q/r]
0 = k/a[(7.74μC+q)√2 + 7.74μC]
or (7.74μC+q)√2 = - 7.74μC
q = -7.74μC[1+1/√2]
Answers
Answered by
bobpursley
correct.
Answered by
sarah
So that gives me q=-13.23. I am told that that is the wrong answer. Please help!
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