the side of an equilateral triangle is 20 cm identical charges of 1.5x 10^-5 C are placed at each of the 3 vertices. If a test charge is placed at the middle of equilateral triangle what force would be acting on the test charge?

20uc

To find the force acting on the test charge at the center of the equilateral triangle, you can use Coulomb's Law. Coulomb's Law states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.

The formula for Coulomb's Law is:

F = k * (q1 * q2) / r^2

Where:
F is the force between the charges.
k is the Coulomb's constant (approximately 9 x 10^9 N m^2/C^2).
q1 and q2 are the magnitudes of the charges.
r is the distance between the charges.

In this case, the charges at each vertex of the equilateral triangle are identical, so q1 and q2 are the same.

Given:
Side of the equilateral triangle = 20 cm
Charge at each vertex = 1.5x10^-5 C

First, let's convert 20 cm to meters:
20 cm = 20/100 = 0.2 m

Now, we can calculate the distance between the charges:
The distance from the center of an equilateral triangle to any vertex is equal to one-third of the height of the equilateral triangle.
The height of an equilateral triangle can be found using the formula:
h = (√3/2) * side

Substituting the values:
h = (√3/2) * 0.2
h ≈ 0.1732 m

So, the distance between the charges (r) is 0.1732 m.

Now, we can calculate the force:
F = k * (q1 * q2) / r^2
F = (9 x 10^9 N m^2/C^2) * (1.5x10^-5 C) * (1.5x10^-5 C) / (0.1732 m)^2

Calculating this expression will give you the force acting on the test charge at the center of the equilateral triangle.