Three charges are at the vertices of an equilateral triangle of 1 cm each side...

Question

Three charges are at the vertices of an equilateral triangle of 1 cm each side. One charge Q is 6700E-6 C , and all the other two charges are 2000E-6 C each. What is the magnitude of the net electric force on the charge Q due to the other two charges?  Blank 1. Fill in the blank, read surrounding text. ___N.  

Answer 1

To solve this problem, first, try to draw out a picture. Let's put the charge Q at the top, and the other two charges are on the bottom. Since all of our charges are positive, we only have repulsive forces. Looking at the charge Q, we have two forces, one from each bottom charge, pointing up and slightly to the left/right. To solve this problem, let's break it down into two parts: finding the magnitude of the force, and summing them together.

Electrostatic Force

Coulomb's Law is,

F_E = k_eq₁q₂/r²

where q₁ and q₂ are two charges, r is the distance of separation between the two points, and k_e is Coulomb's constant, where k_e = 1/(4πε₀) = 8.99x10⁹ Nm²/C².

In our case, because our bottom charges are the same, the force between our top charge Q and either of our bottom charges has the same magnitude, so we only need to calculate this once. We are then given q = 2000x10^-6 C, Q = 6700x10^-6 C, and r = 1 cm. Plugging this in, we get:

F_E = k_e(qQ)/r² = (8.99x10⁹ Nm²/C²)(2000x10^-6 C)(6700x10^-6 C)/((1x10^-2 m)²)

= 1.2047x10⁹ N

= 1.2047 MN

Putting Them Together

To put these two forces together, we need to do a bit of geometry. We have both forces pointing upward, but also pointing a bit to the left and right. Let's break them down into their x- and y-components.

First we find θ, the angle between the force and the vertical axis. This is the same angle as the angle between the left/right arm of the triangle and the vertical axis. Since the triangle is equilateral, bisecting it yields two 30-60-90 triangles, with the top angle being the 30° angle. Thus, θ = 30°.

Next, we find the horizontal and vertical components of the force. The side opposite of θ is the horizontal axis, and the side adjacent is the vertical axis, thus we have that F_x = F_Esin(θ), F_y = F_Ecos(θ).

We choose the force pointing to the right (F₁) to be the positive angle and the left (F₂) to be the negative angle (this choice is arbitrary). Then we have:

(F₁)_x = F_Esin(30°) = (1.2047 MN)(0.5) = 0.60235 MN

(F₁)_y = F_Ecos(30°) = (1.2047 MN)(0.8660) = 1.0433 MN

(F₂)_x = F_Esin(-30°) = (1.2047 MN)(-0.5) = -0.60235 MN

(F₂)_y = F_Ecos(-30°) = (1.2047 MN)(0.8660) = 1.0433 MN

Combining these forces along each axis, we have:

F_x = (F₁)_x + (F₂)_x = 0.60235 MN + -0.60235 MN = 0N

F_y = (F₁)_y + (F₂)_y = 1.0433 MN + 1.0433 MN = 2.0866 MN.

Since our forces cancel in the x, our resultant force is the sum of our y forces, pointing directly up with a magnitude of 2.0866 MN, or 2.0866 meganewtons.