Answer to Question #155640 in Physics for chezca villarin

Question

You have an equilateral triangle, and it happens you place three charges 45, -35, and 77 on each vertex, the equilateral triangle is separated by the distance 61. What is the total net force experience by -35 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 $45\ C$ and $77\ C$, respectively. Let the second charge located at the middle vertex of the equilateral triangle be $-35\ C$. Let, also, the length of the side of the equilateral triangle be $r=61\cdot10^{-15}\ m$ and the side $q_1q_2$ be the $x$-axis.

The force $F_{12}$ directed away from the positive charge $q_1$(toward the negative charge $q_2$). The force $F_{32}$ directed away from the positive charge $q_3$(toward the negative 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:

$$F_{12}=k\dfrac{|q_1q_2|}{r^2},$$ $$F_{12}=9\cdot10^9\ \dfrac{Nm^2}{C^2}\dfrac{|45\ C\cdot(-35\ C)|}{(61\cdot10^{-15}\ m)^2}=3.81\cdot10^{39}\ N,$$ $$F_{32}=k\dfrac{|q_3q_2|}{r^2},$$ $$F_{32}=9\cdot10^9\ \dfrac{Nm^2}{C^2}\dfrac{|77\ C\cdot(-35\ C)|}{(61\cdot10^{-15}\ m)^2}=6.52\cdot10^{39}\ N.$$

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

$$F_x=F_{12x}+F_{32x},$$ $$F_x=3.81\cdot10^{39}\ N+6.52\cdot10^{39}\ N\cdot cos60^{\circ}=7.07\cdot10^{39}\ N,$$ $$F_y=F_{12y}+F_{32y},$$ $$F_y=0+6.52\cdot10^{39}\ N\cdot sin60^{\circ}=5.65\cdot10^{39}\ N.$$

Finally, the net electric force will be:

$$F_{net}=\sqrt{F_x^2+F_y^2},$$ $$F_{net}=\sqrt{(7.07\cdot10^{39}\ N)^2+(5.65\cdot10^{39}\ N)^2}=9.05\cdot10^{39}\ N.$$

Answer:

$F_{net}=9.05\cdot10^{39}\ N.$