Question

Point charges of 6.00 µC and ?2.00 µC are placed 0.100 m apart. (Assume the negative charge is located to the right of the positive charge. Include the sign of the value in your answers.) (a) Where can a third charge be placed so that the net force on it is zero? m to the right of the ?2.00 µC charge (b) What if both charges are positive? m to the right of the 2.00 µC charge

Answer 1

Let us keep the -2? C in between the 6 and the third charge +q ?C.
By doing so, +q ?C will be attracted by the negative charge and repelled by the positive charge.
Force on q? C by the 6C is k [q ?] *[6?] / [0.1 + d] ^2
Force on q? C by the -2? C is k [-2 ?] *[q ?] / d^2

The magnitudes are equal.
k [q ?] *[6?] / [0.10 + d] ^2 = k [2 ?] *[q ?] / d^2
[6] / [0.10 + d] ^2 = [2] / d^2
[0.10 + d] = 1.732 d
d = 0.137 m
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If both were positive, +q ?C. is placed in between them, then as before
[6] / [0.10 - d] ^2 = [2] / d^2
[0.10 - d] = 1.732 d
d = 0.037 m
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