Answer to Question #218724 in Physics for Humanbean

Question

Two point charges are placed as follows: charge q1= -1.50 nC is at y= +6.00 m and charge q2= +3.20 nC is at the origin. What is the total force (magnitude and direction) exerted by these two charges on a negative point charge q3= -5.00 nC located at (2.00 m, -4.00 m)?

Answer 1

Let's first find the magnitude of the force with which the charge $q_1$ acts on charge $q_3$:

$$F_{13}=\dfrac{kq_1q_3}{r_{13}^2}=\dfrac{kq_1q_3}{(\sqrt{r_{12}^2+r_{23}^2})^2} ​ ​ ,$$ $$F_{13}=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot1.5\cdot10^{-9}\ C\cdot5\cdot10^{-9}\ C}{(\sqrt{(6\ m)^2+(\sqrt{(2\ m)^2+(4\ m)^2})^2})^2}=1.205\cdot10^{-9}\ N.$$

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

$$F_{23}=\dfrac{kq_2q_3}{r_{23}^2},$$ $$F_{23}=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot3.2\cdot10^{-9}\ C\cdot5\cdot10^{-9}\ C}{(\sqrt{(2\ m)^2+(4\ m)^2})^2}=7.2\cdot10^{-9}\ N.$$

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

$$F_{13,x}=F_{13}\cdot\dfrac{r_{23}}{r_{13}}=1.205\cdot10^{-9}\ N\cdot\dfrac{\sqrt{(2\ m)^2+(4\ m)^2}}{\sqrt{(6\ m)^2+(\sqrt{(2\ m)^2+(4\ m)^2})^2}},$$ $$F_{13,x}=7.2\cdot10^{-10}\ N.$$ $$F_{13,y}=-F_{13}\cdot\dfrac{r_{12}}{r_{13}}=-1.205\cdot10^{-9}\ N\cdot\dfrac{6\ m}{\sqrt{(6\ m)^2+(\sqrt{(2\ m)^2+(4\ m)^2})^2}},$$ $$F_{13,y}=-9.66\cdot10^{-10}\ N.$$ $$F_{23, x}=-F_{23}\cdot\dfrac{r_{23}}{r_{13}}=-7.2\cdot10^{-9}\ N\cdot\dfrac{\sqrt{(2\ m)^2+(4\ m)^2}}{\sqrt{(6\ m)^2+(\sqrt{(2\ m)^2+(4\ m)^2})^2}},$$ $$F_{23, x}=-4.3\cdot10^{-9}\ N.$$ $$F_{23,y}=F_{23}\cdot\dfrac{r_{12}}{r_{13}}=7.2\cdot10^{-9}\ N\cdot\dfrac{6\ m}{\sqrt{(6\ m)^2+(\sqrt{(2\ m)^2+(4\ m)^2})^2}},$$ $$F_{23,y}=5.77\cdot10^{-9}\ N.$$

Then, we get:

$$F_x=F_{13,x}+F_{23,x}=7.2\cdot10^{-10}\ N+(-4.3\cdot10^{-9}\ N)=-3.58\cdot10^{-9}\ N,$$ $$F_y=F_{13,y}+F_{23,y}=(-9.66\cdot10^{-10}\ N)+5.77\cdot10^{-9}\ N=4.8\cdot10^{-9}\ N.$$

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

$$F=\sqrt{F_x^2+F_y^2},$$ $$F=\sqrt{(-3.58\cdot10^{-9}\ N)^2+(4.8\cdot10^{-9}\ N)^2}=5.98\cdot10^{-9}\ N.$$

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

$$tan\theta=\dfrac{F_y}{F_x},$$ $$\theta=tan^{-1}(\dfrac{F_y}{F_x}),$$ $$\theta=tan^{-1}(\dfrac{4.8\cdot10^{-9}\ N}{-3.58\cdot10^{-9}\ N})=53.3^{\circ}.$$

The total electrostatic force on the charge $q_3$ directed at $53.3^{\circ}$ below the $x$-axis.