The drawing shows a positive point charge + q 1 , a second point charge q 2 that
ID: 1501515 • Letter: T
Question
The drawing shows a positive point charge +q1, a second point charge q2 that may be positive or negative, and a spot labeled P, all on the same straight line. The distance dbetween the two charges is the same as the distance between q1 and the spot P. With q2 present, the magnitude of the net electric field at P is twice what it is when q1 is present alone. Given that q1 = +0.70 ?C, determine the magnitude |q2| when q2 is a) positve and b) negative.
The drawing shows a positive point charge +q1, a second point charge q2 that may be positive or negative, and a spot labeled P, all on the same straight line. The distance dbetween the two charges is the same as the distance between q1 and the spot P. With q2 present, the magnitude of the net electric field at P is twice what it is when q1 is present alone. Given that q1 = +0.70 ?C, determine the magnitude |q2| when q2 is a) positve and b) negative.
Explanation / Answer
At P, the field due to q1 (a positive charge) is negative (that is, to the left).
The field has magnitude k*q1 / d².
The field doubled in magnitude with q2 present must therefore be 2*k*q1/d² or -2*k*q1/d².
In order to be the former, the field due to q2 must be 3*k*q1/d², and since the field due to q2 has magnitude k*q2/(2d)², we have
3*k*q1/d² = k*q2 / 4d² k/d² cancels
3*q1 = q2 / 4
q2 = 12*q1
and since q1 = 0.7 µC, we get
q2 = 8.4 µC
In order to be the latter, the field due to q2 must be -k*q1/d², and since the field due to q2 has magnitude k*q2/(2d)², we have
k*q1/d² = k*q2 / 4d² k/d² cancels
q1 = q2 / 4
q2 = 4*q1 = 4*0.7µC = 2.8 µC
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