Answer to Question #212420 in Electricity and Magnetism for Rocky Valmores

Question

 Three charges are fixed to an xy-coordinate system. A charge of

18µC is on y = 3.0m. A charge of −12µC is at the origin. Lastly, a charge of 45µC is on the x-axis at x = 3.0m. Determine the magnitude and direction of the net electrostatic force on the charge at x = 3.0m

Answer 1

First, we have to determine that the net electrostatic force on the charge at x = 3.0m (with q₃=+45 µC) will be $\overrightarrow{F_{elec}}=\overrightarrow{F_{13}}+\overrightarrow{F_{23}}$

For this problem, we define q₁=+18 µC and q₂=-12 µC as the charges and the distance vectors can be found with the position of all charges on the xy plane as:

$\overrightarrow{r_{13}}=\overrightarrow{r_3}-\overrightarrow{r_1}=(3,0)-(0,3)=(3,-3)$

$r_{13}=\|\overrightarrow{r_{13}}\|=\|(3,-3)\|=\sqrt{(3)^2+(-3)^2}=3\sqrt{2}\,m$

$\widehat{r_{13}}=\frac{\overrightarrow{r_{13}}}{r_{13}}=\frac{(3,-3)}{3\sqrt{2}}=(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$

$\overrightarrow{r_{23}}=\overrightarrow{r_3}-\overrightarrow{r_2}=(3,0)-(0,0)=(3,0)$

$r_{23}=\|\overrightarrow{r_{23}}\|=\|(3,0)\|=\sqrt{(3)^2+(0)^2}=3\,m$

$\widehat{r_{23}}=\frac{\overrightarrow{r_{23}}}{r_{23}}=\frac{(3,0)}{3}=(1,0)$

This defines the forces exerted by each particle on q₃ as:

$\overrightarrow{F_{13}}=k\dfrac{q_1q_3}{{r_{13}}^2}\,\widehat{{r_{13}}}$

$\overrightarrow{F_{13}}=(8.988\times10^9\frac{Nm^2}{C^2})\frac{(18\times10^{-6}C)(45\times10^{-6}C)}{(3\sqrt{2}\,m)^2}(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$

$\overrightarrow{F_{13}}=(0.40446\,N)(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})\approxeq(0.2860,-0.2860)N$

$\overrightarrow{F_{23}}=k\dfrac{q_2q_3}{{r_{23}}^2}\,\widehat{{r_{23}}}$

$\overrightarrow{F_{23}}=(8.988\times10^9\frac{Nm^2}{C^2})\frac{(45\times10^{-6}C)(-12\times10^{-6}C)}{(3\,m)^2}(1,0)$

$\overrightarrow{F_{23}}=(-0.53928\,N)(1,0)\approxeq(-0.53928,0)N$

Then the total force will be the sum:

$\overrightarrow{F_{elec}}=\overrightarrow{F_{13}}+\overrightarrow{F_{23}} \approxeq (0.2860,-0.2860)N+(-0.53928,0)N$

$\overrightarrow{F_{elec}}=\overrightarrow{F_{1/3}}+\overrightarrow{F_{2/3}} \approxeq (-0.25328,-0.2860)N$

Thus the magnitude of this new vector is:

$F_{elec}=\|\overrightarrow{F_{elec}}\| \approxeq\|(-0.25328,-0.2860)\|N\approxeq[\sqrt{(-0.25328)^2+(-0.2860)^2}]N$

$F_{elec}=\|\overrightarrow{F_{elec}}\| \approxeq 0.3820\,N$

To find the angle or orientation we use a trigonometric function:

$\tan \theta =\frac{y}{x}=\frac{-0.2860}{-0.25328}=1.1291 \implies \theta=\arctan(1.1291)=48.47°$

That last means the vector is located at the third quadrant on $\theta'=228.47°$

In conclusion, the magnitude of the force on x=3.0 m is F=0.3820 N and the orientation is near to $\theta'=228.47°$ with respect to the x-axis thus the force vector is located on the third quadrant.

Reference:

• Young, H. D., & Freedman, R. A. (2015). University Physics with Modern Physics and Mastering Physics (p. 1632). Academic Imports Sweden AB.