3 equal charges each of of 2*0.000001C are fixed at three corners of an equilateral triangle ofsides 5 cm.find the column force experienced by one of the charge due to the other two

See post for THOMAS123 on Jan 12.

It will give you the framework

To find the electric force experienced by one of the charges due to the other two charges, we can use Coulomb's law. Coulomb's law states that the electric force (F) between two point charges is directly proportional to the product of their magnitudes (Q1 and Q2) and inversely proportional to the square of the distance (r) between them.

The formula for the electric force is:

F = (k * Q1 * Q2) / r^2

Where F is the electric force, Q1 and Q2 are the magnitudes of the charges, r is the distance between the charges, and k is the Coulomb's constant.

Given that all three charges are equal to 2 * 0.000001C (convert to scientific notation: 2 x 10^-6C), and the side length of the equilateral triangle is 5 cm, we can calculate the force by considering one of the charges as the source (Q1) and the other two charges as the objects (Q2).

Let's assume charge A is the source charge and charges B and C are the object charges. So, we need to calculate the force experienced by charge A due to charges B and C.

Step 1: Calculate the distance between the charges.
The distance between charges A and B is equal to the side length of the equilateral triangle. So, rAB = 5 cm.
The distance between charges A and C is also equal to the side length of the equilateral triangle. So, rAC = 5 cm.

Step 2: Convert the distance from cm to meters.
To use Coulomb's law, we need to convert the distance from centimeters to meters. Since 1 cm = 0.01 m, the distances become:
rAB = 0.05 m
rAC = 0.05 m

Step 3: Calculate the electric force experienced by charge A due to charges B and C using Coulomb's law.
Using the formula for electric force, we have:
FAB = (k * Q1 * Q2) / rAB^2 (force experienced by charge A due to charge B)
FAC = (k * Q1 * Q2) / rAC^2 (force experienced by charge A due to charge C)

Substituting the given values, we get:
FAB = (9 x 10^9 Nm^2/C^2 * (2 x 10^-6 C)^2) / (0.05 m)^2
FAC = (9 x 10^9 Nm^2/C^2 * (2 x 10^-6 C)^2) / (0.05 m)^2

Calculating further, we get:
FAB = 7.2 x 10^-5 N
FAC = 7.2 x 10^-5 N

Therefore, the electric force experienced by one of the charges due to the other two charges is approximately 7.2 x 10^-5 Newtons.