Question

1. a point charge ( q = - 12 mu c ), and two other charges, ( q _ { 1 } ) and ( q _ { 2 } ), are placed as shown. the electric force components on charge ( q ) are ( f _ { x } = 0.005 mathrm { n } ) and ( f _ { y } = - 0.003 mathrm { n } ) as shown in the figure. charge ( q _ { 2 } ), in ( mathrm { nc } ), is closest to:

a. ( + 480 mathrm { nc } )
b. ( - 480 mathrm { nc } )
c. ( + 640 mathrm { nc } )
d. ( - 640 mathrm { nc } )

Explanation:

Step1: Analyze the force components

The \(x -\)component of the force \(F_x\) is due to charge \(q_1\), and the \(y -\)component of the force \(F_y\) is due to charge \(q_2\).
We use Coulomb's law \(F=\frac{k|q_1q_2|}{r^{2}}\), where \(k = 9\times10^{9}\ N\cdot m^{2}/C^{2}\).
For the \(y -\)direction, \(F_y=\frac{k|Qq_2|}{r_{y}^{2}}\) (since \(F_y\) is the force on \(Q\) due to \(q_2\)).
We can re - arrange the formula for \(q_2\): \(|q_2|=\frac{|F_y|r_{y}^{2}}{k|Q|}\)

Step2: Substitute the values

Given \(F_y=- 0.003\ N\) (we take the magnitude \(|F_y| = 0.003\ N\)), \(r_y = 4.8\ m\), \(Q=-12\times10^{-6}\ C\) (so \(|Q| = 12\times10^{-6}\ C\)) and \(k = 9\times10^{9}\ N\cdot m^{2}/C^{2}\)

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Since \(F_y\) is in the negative \(y -\)direction and \(Q\) is negative, \(q_2\) is positive (because the force between \(Q\) and \(q_2\) is attractive, so \(q_2\) has the opposite sign of \(Q\))

Answer:

A. \(+ 480\ nC\) (There is a miscalculation in the above step - by - step. Let's re - calculate:

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Since \(F_y<0\) (attractive force between \(Q=-12\ \mu C\) and \(q_2\)), \(q_2>0\))
A. \(+480\ nC\)