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Question

Three point charges are arranged on a line. Charge q3=+5.00nCq _ { 3 } = + 5.00 n C and is at the origin. Charge q2=3.00nCq _ { 2 } = - 3.00 n C and is at x = +4.00 cm. Charge q1q_1 is at x = +2.00 cm. What is q1q_1 (magnitude and sign) if the net force on q3q_3 is zero?

Solution

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Answered 1 year ago
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Given

We are given three charges separated with a distance as shown below. Where q2q_{2} = - 3.0 nC and q3q_{3} = +5.0 nC. The charges q1q_{1} and q2q_{2} exerted a force on charge q3q_{3} where the net force that acts on q3q_{3} is zero.

Solution

In this problem q3q_{3} is affected by two forces F1on3F_{1 \,\text{on}\, 3} and F2on3F_{2 \,\text{on}\, 3} and the net force is

Fnet=F1on3+F2on3F_{\text{net}} = F_{1 \,\text{on}\, 3} + F_{2 \,\text{on}\, 3}

Also, we are given the charge q1q_{1} in the positive direction of xx which means there is a repulsive force between q3q_{3} and q1q_{1} and we can predict that the charge q1q_{1} is positive\textbf{positive}. Also, as the net force is zero, this means

F1on3=F2on3F_{1 \,\text{on}\, 3} = F_{2 \,\text{on}\, 3}

Now we can apply Coulomb's law to find the magnitude of q1q_{1} by the next steps

F1on3=F2on3kq1q3r12=kq2q3r22q1r12=q2r22q1=q2(r12r22)\begin{gather*} F_{1 \,\text{on}\, 3} = F_{2 \,\text{on}\, 3} \\ \cancel k \dfrac{|q_{1} \cancel q_{3}|}{r_{1}^{2}} = \cancel k \dfrac{|q_{2}\cancel q_{3}|}{r_{2}^{2}}\\ \dfrac{|q_{1}|}{r_{1}^{2}} = \dfrac{|q_{2}|}{r_{2}^{2}} \tag{Solve for $q_{1}$}\\ |q_{1}| = |q_{2}| \left(\dfrac{r_{1}^{2}}{r_{2}^{2}}\right) \tag{1} \end{gather*}

Now we can plug our values for q2,r1q_{2}, r_{1} and r2r_{2} into equation (1) to get q1q_{1}

q1=q2(r12r22)=3.0nC((2.0cm)2(4.0cm)2)=+0.75nC\begin{align*} |q_{1}| &= |q_{2}| \left(\dfrac{r_{1}^{2}}{r_{2}^{2}}\right)\\ &= |-3.0 \,\text{nC}| \left(\dfrac{(2.0 \,\text{cm})^{2}}{(4.0 \,\text{cm})^{2}}\right)\\ &= \boxed{+ 0.75 \,\text{nC}} \end{align*}

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