At a particular moment, one negative and two positive charges are located in the xy plane as ...

Question:

At a particular moment, one negative and two positive charges are located in the xy plane as charge {eq}Q_1 = 2\ \mu C {/eq} at {eq}(x_1 = 0, y_1 = 4\ cm), Q_2 = 9\ \mu C {/eq} at origin, and {eq}Q_3 = -7\ \mu C {/eq} at {eq}(x_3 = 3\ cm, y_3 = 0) {/eq}. What is the electric field at location {eq}A (x_4 = 3\ cm, y_4 = 4\ cm) {/eq}, due to all three charges in vector form?

Electric Field on a Point Due to System of Charges:

In the presence of several point charges, the net electric field at a certain point in space is basically the sum of the components of the electric field due to each charge. This also applies to the net electric force exerted on a point charge that is placed in the location where the electric field is measured.

Answer and Explanation:

Given:

{eq}\begin{align} Q_1&=2.0\times 10^{-6}\,\rm C\,\,@\,\,(0,0.04\,\rm m)\\[0.2cm] Q_2&=9.0\times 10^{-6}\,\rm C\,\,@\,\,(0,0)\\[0.2cm] Q_3&=-7.0\times 10^{-6}\,\rm C\,\,@\,\,(0.03\,\rm m, 0)\\[0.2cm] \end{align} {/eq}

The electric field at {eq}(0.03\,\rm m, 0.04\,\rm m) {/eq} due to each charge would be given by

$$\begin{align} \vec{E_1}&=\dfrac{kQ_1}{r_1^2}\,\hat{x}\\[0.2cm] \vec{E_2}&=\dfrac{kQ_2}{r_2^2}\cos{\theta_2}\,\hat{x}+\dfrac{kQ_2}{r_2^2}\sin{\theta_2}\,\hat{y}\\[0.2cm] \vec{E_3}&=\dfrac{kQ_3}{r_3^2}\,\hat{y}\\[0.2cm] \end{align} $$

where {eq}k\,(8.99\times 10^9\,\rm N\cdot \rm m^2/\rm C^2) {/eq} is the electric constant, {eq}r_1, r_2 {/eq}, and {eq}r_3 {/eq} are the distance from {eq}Q_1, Q_2 {/eq}, and {eq}Q_3 {/eq} to point {eq}(0.03\,\rm m, 0.04\,\rm m) {/eq}, respectively, and {eq}\theta_2 {/eq} is the angle between the electric field due to {eq}Q_2 {/eq} from the positive x-axis.

The values of the distances are given by

$$\begin{align} r_1&=\sqrt{(0-0.03\,\rm m)^2+(0.04\,\rm m-0.04\,\rm m)^2}=0.03\,\rm m\\[0.2cm] r_2&=\sqrt{(0.03\,\rm m)^2+(0.04\,\rm m)^2}=0.05\,\rm m\\[0.2cm] r_3&=\sqrt{(0.03\,\rm m-0.03\,\rm m)^2+(0.04\,\rm m-0)^2}=0.04\,\rm m\\[0.2cm] \theta_2&=\tan^{-1} \dfrac{0.04\,\rm m}{0.03\,\rm m}=53.13^\circ\\[0.2cm] \end{align} $$

By substituting the known values, we will have

$$\begin{align} \vec{E_1}&=\dfrac{(8.99\times 10^9\,\rm N\cdot \rm m^2/\rm C^2)(2.0\times 10^{-6}\,\rm C)}{(0.03\,\rm m)^2}\,\hat{x}\\[0.2cm] &\approx 2.00\times 10^7\,\rm N/\rm C\,\hat{x}\\[0.2cm]\\[0.2cm] \vec{E_2}&=\dfrac{(8.99\times 10^9\,\rm N\cdot \rm m^2/\rm C^2)(9.0\times 10^{-6}\,\rm C)}{(0.05\,\rm m)^2}\cos{(53.13^\circ)}\,\hat{x}+\dfrac{(8.99\times 10^9\,\rm N\cdot \rm m^2/\rm C^2)(9.0\times 10^{-6}\,\rm C)}{(0.05\,\rm m)^2}\sin{(53.13^\circ)}\,\hat{y}\\[0.2cm] &\approx 1.94\times 10^7\,\rm N/\rm C\,\hat{x} + 2.59\times 10^7\,\rm N/\rm C\,\hat{y}\\[0.2cm]\\[0.2cm] \vec{E_3}&=\dfrac{(8.99\times 10^9\,\rm N\cdot \rm m^2/\rm C^2)(-7.0\times 10^{-6}\,\rm C)}{(0.04\,\rm m)^2}\,\hat{y}\\[0.2cm] &\approx -3.93\times 10^7\,\rm N/\rm C\,\hat{y}\\[0.2cm] \end{align} $$

Therefore, the net electric field at {eq}(0.03\,\rm m, 0.04\,\rm m) {/eq} due to the three charges would be

$$\begin{align} \vec{E_{net}}&=\vec{E_1}+\vec{E_2}+\vec{E_3}\\[0.2cm] &\approx (2.00\times 10^7\,\rm N/\rm C\,\hat{x}) + (1.94\times 10^7\,\rm N/\rm C\,\hat{x} + 2.59\times 10^7\,\rm N/\rm C\,\hat{y}) + (-3.93\times 10^7\,\rm N/\rm C\,\hat{y})\\[0.2cm] &\color{red}{\approx 3.94\times 10^7\,\rm N/\rm C\,\hat{x} - 1.34\times 10^7\,\rm N/\rm C\,\hat{y} } \end{align} $$


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