Question

A negative charge of -0.550 μC exerts an upward 0.900-N force on an unknown charge that is located 0.300 m directly below the first charge. Part A What is the value of the unknown charge (magnitude and sign)? Express your answer with the appropriate units.

Answer 1

Answer:

$1.6381\times 10^{-5}\ C$

Explanation:

F = Force between charge = 0.9 N

$q_1$ = Charge on particle = -0.55 μC

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Electric Force is given by

$F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow 0.9=\dfrac{8.99\times 10^9\times -0.55q_2}{0.3^2}\\\Rightarrow q_2=\dfrac{0.9\times 0.3^2}{8.99\times 10^9\times 0.55\times 10^{-6}}\\\Rightarrow q_2=+1.6381\times 10^{-5}\ C$

As there is a force of attraction the other charge will have the opposite sign

The magnitude and sign of the charge is $+1.6381\times 10^{-5}\ C$