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 -2u C in between the 6 and the third charge +q uC.

By doing so, +q uC will be attracted by the negative charge and repelled by the positive charge.

Force on q uC by the 6 uC is k [q u] *[5u] / [0.1 + d] ^2

Force on q uC by the -2 uC is k [-2 u] *[q u] / d^2

The magnitudes are equal.

k [q u] *[6u] / [0.1 + d] ^2 = k [2 u] *[q u] / d^2

[6] / [0.1 + d] ^2 = [2] / d^2

[0.1 + d] = 3 d

d = 0.05 m

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If both were positive, +q uC. is placed in between them, then as before

[56 / [0.1 - d] ^2 = [2] / d^2

[0.1 - d] = 3 d

d = 0.025m

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