To solve this problem, we need to calculate the electrostatic force on the charge $2q$ due to the other charges $2q$ and $-q$, using Coulomb's Law.

Given Data

• Charge $q = 10^{-8} \, \text{C}$
• Distance $a = 3 \, \text{mm} = 0.003 \, \text{m}$
• Charges: $2q$ at two vertices and $-q$ at the top vertex of the triangle

Steps

1. Calculate the Force between $2q$ and $-q$: The force $F$ between two charges is given by: $F = \frac{k |q_1 q_2|}{r^2}$ where $k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2$ (Coulomb's constant), $q_1 = 2q$, $q_2 = -q$, and $r = a$.

   Plugging in the values: $F_{2q, -q} = \frac{8.99 \times 10^9 \times |2 \times 10^{-8} \times -10^{-8}|}{(0.003)^2}$ Simplify to find the numerical value of this force.

2. Calculate the Force between $2q$ and the other $2q$: Since they are both $2q$, the force is repulsive. Using the same formula, $F = \frac{k |q_1 q_2|}{r^2}$, where $q_1 = 2q$ and $q_2 = 2q$, and $r = a$: $F_{2q, 2q} = \frac{8.99 \times 10^9 \times |2 \times 10^{-8} \times 2 \times 10^{-8}|}{(0.003)^2}$

3. Resolve Forces into Components: Since the triangle is equilateral, the angle between the lines connecting any two charges is $60^\circ$. Decompose the forces from each charge into $x$- and $y$-components to find the resultant force on the $2q$ charge on the left.

4. Calculate the Resultant Force: Sum the $x$- and $y$-components of the forces to get the total force on the $2q$ charge at the bottom left.

Would you like me to go through each calculation step-by-step, or do you have any specific question?

Related Questions:

1. How would the force change if $a$ were doubled?
2. What if the charge $-q$ were replaced by $+q$?
3. How do you calculate the net force when charges are arranged in different geometries?
4. What role does Coulomb’s constant play in force magnitude?
5. How would the direction of the force change if all charges were positive?

Tip: Use vector decomposition for forces in equilateral triangle setups to simplify the calculation.