Question

Point charges of 6.75 µC and

−4.00 µC

are placed 0.300 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

−4.00 µC

charge

(b) What if both charges are positive?
m to the right of the  4.00 µC charge

Answer 1

(a) Let us consider that -4 μC is between the 6.75 μC and the third charge +q μC.

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

Now -  
Force on q μC by the 6.75 μC is k [q μ] *[5μ] / [0.30 + d] ^2

And, force on qμ C by the -4.0 μ C is k [-4 μ] *[q μ] / d^2

The magnitudes are equal, therefore -

k [q μ] *[6.75μ] / [0.30 + d] ^2 = k [4 μ] *[q μ] / d^2

=> [6.75] / [0.30 + d] ^2 = [4] / d^2

take square root of both sides -

=> 2.60 / [0.30 + d] = 2/d

=> 2.60 * d = 2 * (0.30 + d)

=> 0.60 * d = 0.60

=> d = 1.0 m

So, the third charge is located at 1.0 m from -4.0 µC charge.

(b) If both were positive, +q μC. is placed in between them, then as before
[6.75] / [0.30 - d] ^2 = [4] / d^2

take square-root of both sides -

=> 2.60 / [0.30 - d] = 2 / d

=> 2.60*d = 2(0.30 - d)

=> 4.60*d = 0.60

=> d = 0.60 / 4.60 = 0.13 m
So, the third charge is 0.13 m from 4.0 µC charge.