Answer to Question #211505 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

Let the charge of 18 µC be $q_1$, the charge of −12 µC be $q_2$ and the charge of 45 µC be $q_3$. Let's first find the magnitude of the force with which the charge $q_1$ acts on charge $q_3$:

$$F_{13}=\dfrac{kq_1q_3}{r_{13}^2}=\dfrac{kq_1q_3}{(\sqrt{r_{12}^2+r_{23}^2})^2},$$ $$F_{13}=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot18\cdot10^{-6}\ C\cdot45\cdot10^{-6}\ C}{(\sqrt{(3\ m)^2+(3\ m)^2})^2}=0.405\ N.$$

Similarly, let's find the magnitude of the force with which the charge $q_2$ acts on charge $q_3$:

$$F_{23}=\dfrac{kq_2q_3}{r_{23}^2},$$ $$F_{23}=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot12\cdot10^{-6}\ C\cdot45\cdot10^{-6}\ C}{(3\ m)^2}=0.54\ N.$$

Let's find the projections of forces $F_{13}$ and $F_{23}$ on axis $x$ and $y$:

$$F_{13,x}=F_{13}cos45^{\circ}=0.405\ N\cdot cos45^{\circ}= 0.286\ N,$$ $$F_{13,y}=F_{13}sin45^{\circ}=0.405\ N\cdot sin45^{\circ}= -0.286\ N,$$ $$F_{23,x}=-0.54\ N,$$ $$F_{23,y}=0.$$

Then, we get:

$$F_x=F_{13,x}+F_{23,x}=0.286\ N+(-0.54\ N)=-0.254\ N,$$ $$F_y=F_{13,y}=-0.286\ N.$$

We can find the net electrostatic force on the charge $q_3$ from the Pythagorean theorem:

$$F=\sqrt{F_x^2+F_y^2},$$ $$F=\sqrt{(-0.254\ N)^2+(-0.286\ N)^2}=0.382\ N.$$

We can find the direction of the net electrostatic force on the charge $q_3$ from the geometry:

$$tan\theta=\dfrac{F_y}{F_x},$$ $$\theta=tan^{-1}(\dfrac{F_y}{F_x}),$$ $$\theta=tan^{-1}(\dfrac{-0.286\ N}{-0.254\ N})=48.4^{\circ}.$$

The net electrostatic force on the charge $q_3$ directed at $48.4^{\circ}$ below the $x$-axis.