A point charge of $3.0 \mu \mathrm{C}$ is located on the $y$-axis at $y=4.0 \mathrm{~m}$, and another point charge of $6.0 \mu \mathrm{C}$ is located on the $\mathrm{x}$-axis at $\mathrm{x}=5.0 \mathrm{~m}$. What is the direction (ccw from the $\mathrm{x}$-axis) of the electric field at $(5.0 \mathrm{~m}, 4.0 \mathrm{~m}$ )?

Question

Quizplus · Accepted Answer

The electric field due to the charge at $y=4.0 \mathrm{~m}$ is in the $-y$ direction, and the electric field due to the charge at $x=5.0 \mathrm{~m}$ is in the $+x$ direction. The net electric field is therefore in the first quadrant, and its direction is given by$$\tan \theta = \frac{E_y}{E_x} = \frac{k\frac{3\mu \mathrm{C}}{(4.0 \mathrm{~m})^2}}{k\frac{6\mu \mathrm{C}}{(5.0 \mathrm{~m})^2}} = \frac{3}{4}$$$$\theta = \tan^{-1} \left(\frac{3}{4}\right) = 37^\circ$$The direction of the electric field is therefore $37^\circ$ ccw from the $\mathrm{x}$-axis.

A Point Charge Of 3.0μC3.0 \mu \mathrm{C}3.0μC Is Located on The yyy

A Point Charge Of $3.0 \mu \mathrm{C}$ Is Located on The $y$