Question

Particles q₁ = +18.1 µC, q2 = -11.2 μC, and 93 = +5.67 μC are in a line. Particles q₁ and q2 are separated by 0.280 m and particles q2 and q3 are separated by 0.350 m. What is the net force on particle q₂? ​

Answer 1

Answer:

$\vec F_{net \ on \ q_2}=15.9637N}} \ \text{at 180\textdegree (or to the left/negative x-axis)}$

Explanation:

Using Coulomb's law to answer this question.

$\boxed{\left\begin{array}{ccc}\text{\underline{Coloumb's Law:}}\\\\\vec F_e=\frac{k_eq_1q_2}{r^2} \cdot \hat r \end{array}\right}$

• $k_e$ is Coulomb's constant ($8.99 \times10^9\frac{Nm^2}{C^2}$)
• $\hat r$ is a direction vector that points towards the charge you are calculating the force on

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Given:

$q_1=18.1 \ \mu C \rightarrow 18.1 \times 10 ^{-6} \ C\\r_{2_{1}}=0.280 \ m\\q_2=-11.2 \ \mu C \rightarrow -11.2 \times 10 ^{-6} \ C\\q_3=5.67 \ \mu C \rightarrow 5.67 \times 10 ^{-6}\ C\\r_{2_{3}}=0.350 \ m$

Find:

$|| \vec F_{net \ on \ q_{2}}||= \ ?? \ N$

**Assuming q_2 is the center of a coordinate system.

(1) - Find the force on q_2 exerted by q_1

$\vec F_{2_{1}}=\frac{k_eq_2q_1}{(r_{2_{1}})^2} \cdot \hat r_{2_{1}}\\\\\Longrightarrow \vec F_{2_{1}}=\frac{(8.99 \times 10^{9})(-11.2 \times 10^{-6})(18.1 \times 10^{-6})}{(0.280)^2} \cdot \frac{ < 0.280,0 > }{\sqrt{(0.280)^2+(0)^2} } \\\\\Longrightarrow \vec F_{2_{1}}=-23.2456 \cdot < 1,0 > \\\\\therefore \boxed{\vec F_{2_{1}}= < -23.2456,0 > N}$

(2) - Find the force on q_2 exerted by q_3

$\vec F_{2_{1}}=\frac{k_eq_2q_3}{(r_{2_{3}})^2} \cdot \hat r_{2_{3}}\\\\\Longrightarrow \vec F_{2_{3}}=\frac{(8.99 \times 10^{9})(-11.2 \times 10^{-6})(5.67 \times 10^{-6})}{(0.280)^2} \cdot \frac{ < - 0.350,0 > }{\sqrt{(-0.350)^2+(0)^2} } \\\\\Longrightarrow \vec F_{2_{3}}=-7.2819 \cdot < -1,0 > \\\\\therefore \boxed{\vec F_{2_{3}}= < 7.2819,0 > N}$

(3) - Find the net charge on q_2

$\vec F_{net \ on \ q_2}=\vec F_{2_{1}}+\vec F_{2_{3}}\\\\\text{Recall that} \ \vec F_{2_{1}}= < -23.2456,0 > N \ \text{and} \ \vec F_{2_{3}}= < 7.2819,0 > N\\\\\Longrightarrow \vec F_{net \ on \ q_2}= < -23.2456,0 > + < 7.2819,0 > \\\\\therefore \boxed{\vec F_{net \ on \ q_2}= < -15.9637,0 > N}\\\\\Longrightarrow ||\vec F_{net \ on \ q_2}||=\sqrt{(-15.9637)^2+(0)^2} \\\\\therefore \boxed{\boxed{||\vec F_{net \ on \ q_2}||=15.9637N}} \ \text{at 180\textdegree (or to the left/negative x-axis)}$