Answer to Question #155848 in Electric Circuits for Audrey

Question

1. You have an equilateral triangle, and it happens you place three charges 16C 18C 17C on each vertex, the equilateral triangle is separated by the distance r. What is the total net force experience by q2 due to other two charges if the charge is place at the middle vertex of the triangle?

Answer 1

Let the first and third charges located at the base of the equilateral triangle be $16\ C$ and $17\ C$, respectively. Let the second charge located at the middle vertex of the equilateral triangle be $18\ C$. Let, also, the length of the side of the equilateral triangle be $r=50\cdot10^{-15}\ m$ and the side $q_1q_2$ be the $x$-axis.

The force $F_{12}$ directed away from the positive charge $q_1$(toward the positive charge $q_2$). The force $F_{32}$ directed away from the positive charge $q_3$(toward the positive charge $q_2$). It is obviously, that the net electric force on charge $q_2$ due to charges $q_1$ and $q_3$ is the vector sum of forces $F_{12}$ and $F_{32}$.

Let’s find the magnitudes of these forces:

$$F_{12}=k\dfrac{|q_1q_2|}{r^2},$$ $$F_{12}=9\cdot10^9\ \dfrac{Nm^2}{C^2}\dfrac{|16\ C\cdot18\ C|}{(50\cdot10^{-15}\ m)^2}=1.04\cdot10^{39}\ N,$$ $$F_{32}=k\dfrac{|q_3q_2|}{r^2},$$ $$F_{32}=9\cdot10^9\ \dfrac{Nm^2}{C^2}\dfrac{|17\ C\cdot18\ C|}{(50\cdot10^{-15}\ m)^2}=1.1\cdot10^{39}\ N.$$

Then, we can find the projections of forces $F_{12}$ and $F_{32}$ on axis $x$ and $y$:

$$F_x=F_{12x}+F_{32x},$$ $$F_x=1.04\cdot10^{39}\ N+1.1\cdot10^{39}\ N\cdot cos60^{\circ}=1.6\cdot10^{39}\ N,$$ $$F_y=F_{12y}+F_{32y},$$ $$F_y=0+1.1\cdot10^{39}\ N\cdot sin60^{\circ}=9.53\cdot10^{38}\ N.$$

Finally, the net electric force will be:

$$F_{net}=\sqrt{F_x^2+F_y^2},$$ $$F_{net}=\sqrt{(1.6\cdot10^{39}\ N)^2+(9.53\cdot10^{38}\ N)^2}=1.86\cdot10^{39}\ N.$$

Answer:

$F_{net}=1.86\cdot10^{39}\ N.$