Question

two charges +3uC and +12uC are fixed 1 m apart with the second one to the right. find the magnitude and direction of the net force on a -2-nC charge when placed at the following locations:

(a) halfway between the two

(b) half a m to the left of the +3uC charge

(c) half a m above the +12uC charge in a direction perpendicular to the line joining the two fixed charges.

Answer 1

(a) Let $q_1=3\ \mu C$, $q_2=-2\ nC$ and $q_3=12\ \mu C$.Then, we can find the net force on a −2nC charge as follows:

$$F_{net}=F_{32}-F_{12},$$ $$F_{net}=\dfrac{k}{r^2}(|q_2q_3|-|q_1q_2|),$$

$F_{net}=\dfrac{8.99\cdot10^9\ \dfrac{Nm^2}{C^2}}{(0.5\ m)^2}(|(-2\cdot10^{-9}\ C)\cdot12\cdot10^{-6}\ C|-|3\cdot10^{-6}\ C\cdot(-2\cdot10^{-9}\ C)|)=6.47\cdot10^{-4}\ N.$

The sign plus means that the net force directed to the right.

(b) Let $q_1=-2\ nC$, $q_2=3\ \mu C$ and $q_3=12\ \mu C$.Then, we can find the net force on a −2nC charge as follows:

$$F_{net}=F_{21}+F_{31},$$ $$F_{net}=k(\dfrac{|q_1q_2|}{r_{21}^2}+\dfrac{|q_1q_3|}{r_{31}^2}),$$

$F_{net}=8.99\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot(\dfrac{|(-2\cdot10^{-9}\ C)\cdot3\cdot10^{-6}\ C|}{(0.5\ m)^2}+\dfrac{|(-2\cdot10^{-9}\ C)\cdot12\cdot10^{-6}\ C|}{(1.5\ m)^2})=3.12\cdot10^{-4}\ N.$

The sign plus means that the net force directed to the right.

(c) Let $q_1=3\ \mu C$, $q_2=12\ \mu C$ and $q_3=-2\ nC$. Let's first find the $x$- and $y$-components of the electric force:

$$F_{13,x}=-k\cdot\dfrac{|q_1q_3|}{r_{13}^2}\cdot\dfrac{r_{23}}{r_{13}},$$

$F_{13,x}=-8.99\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot\dfrac{|3\cdot10^{-6}\ C\cdot(-2\cdot10^{-9}\ C)|}{(\sqrt{(1\ m)^2+(0.5\ m)^2})^2}\cdot\dfrac{0.5\ m}{\sqrt{(1\ m)^2+(0.5\ m)^2}}=-1.93\cdot10^{-5}\ N,$

$$F_{23,x}=0\ N,$$ $$F_{13,y}=-k\cdot\dfrac{|q_1q_3|}{r_{13}^2}\cdot\dfrac{r_{12}}{r_{13}},$$

$F_{13,y}=-8.99\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot\dfrac{|3\cdot10^{-6}\ C\cdot(-2\cdot10^{-9}\ C)|}{(\sqrt{(1\ m)^2+(0.5\ m)^2})^2}\cdot\dfrac{1.0\ m}{\sqrt{(1\ m)^2+(0.5\ m)^2}}=-3.86\cdot10^{-5}\ N,$

$$F_{23,y}=-k\dfrac{|q_2q_3|}{r_{23}^2},$$

$F_{23,y}=-8.99\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot\dfrac{|12\cdot10^{-6}\ C\cdot(-2\cdot10^{-9}\ C)|}{(0.5\ m)^2}=-8.63\cdot10^{-4}\ N,$

$$F_x=F_{13,x}=-1.93\cdot10^{-5}\ N,$$ $$F_y=F_{13,y}+F_{23,y},$$ $$F_y=-3.86\cdot10^{-5}\ N+(-8.63\cdot10^{-4}\ N)=-9.02\cdot10^{-4}\ N.$$

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

$$F=\sqrt{F_x^2+F_y^2},$$ $$F=\sqrt{(-1.93\cdot10^{-5}\ N)^2+(-9.02\cdot10^{-4}\ N)^2}=9.02\cdot10^{-4}\ N.$$

We can find the direction of the net electric force from the geometry:

$$\theta=cos^{-1}\dfrac{F_x}{F},$$ $$\theta=cos^{-1}\dfrac{-1.93\cdot10^{-5}\ N}{9.02\cdot10^{-4}\ N}=91.2^{\circ}.$$

The net electric force on charge $q_3$directed $91.2^{\circ}$ below $x$-axis.