# Calculating force on the charges placed in the corners of a triangle

• mmoadi
In summary, three charges are placed in the corners of an equilateral triangle with side length of 1 m. The charges are sized e1, e2 = −2e1, and e3 = 3e1, where e1 = 10^-6 As. The force on the first charge is calculated to be 2.39 N in the upward and rightward direction. The force on the second charge is found to be 0.87 N upward and 0.9 N leftward, while the force on the third charge is 1.35 N leftward and 2.34 N upward.
mmoadi

## Homework Statement

We place three charges in the corners of the equilateral triangle; the size of the sides is 1 m. Charges are of sizes: e1, e2 = −2e1, and e3 = 3e1, where e1 = 10 ^-6 As. What is the force on the first (second, third) charge and in which direction are they showing?

http://item.slide.com/r/1/20/i/uNrFtWB4tT_sS7zkuV1ExJS_gf3980kJ/
http://item.slide.com/r/1/130/i/1G12EjLq5D8Lo4lU8yYKxmOQfVTQfoSl/
http://item.slide.com/r/1/142/i/dpHAZtcg7j-o-U_uDNFiH60YH-Do8Dwr/

## Homework Equations

F= k[q(1)q(2)/ d²]

## The Attempt at a Solution

Calculating for e(1):

d= 1 m

F(1,2)= k[q(1)q(2)/ d²] → F(1,2)= 1.8 N

F(1,2-y)= -F(1,2) sin 60º= - 1.56 N
F(1,2-x)= -F(1,2) cos 60º= - 0.4 N

F(1,3)= k[q(1)q(3)/ d²] → F(1,3)= 2.7 N

F(1,3-y)= f(1,3) sin 60º= 2.34 N
F(1,3-x)= -F(1,3) cos 60º= - 1.35 N

F(1-x)= -2.25 N
F(1-y)= F(1,2-y) + F(1,3-y)= 0.78 N

F(1)= sqrt[F(1-x)² + F(1-y)²]= 2.39 N

Are my calculations for e(1) correct?
Thank you for helping!

Last edited by a moderator:
Have't looked at them all but cos 60=0.5 so not sure where F(1,2-x) came from.

mmoadi said:

## Homework Statement

F(1,2-x)= -F(1,2) cos 60º= - 0.4 N

My mistake, it was a typo. here is the correct calculation as I made it:

F(1,2-x)= -F(1,2) cos 60º= - 0.9 N

Now are my calculations correct for e(1)?
Thank you for helping!

My results are different:

Here is how I did force of particle 2 on 1: k= 9.0 E9

total force (2 on 1) = K (1E-6)(2E-6)/1m^2=18 E-3. Since this is attractive force, it is directed upward and rightward.

Fx = F(2-1) cos 60= 9 E-3N
Fy = F(2-1) sin 60 = 0.87 * 18E-3= 15.6 E-3

Do the same for 3 on 1, and then add x components, then Y components and use Pythags formula for total. Direction can be found using tan(angle)=sum y/ sum x

Thank you for helping, I think I understand now

## What is the formula for calculating the force on charges placed in the corners of a triangle?

The formula for calculating the force on charges placed in the corners of a triangle is F = k(q1q2)/r^2, where F is the force, k is the Coulomb's constant, q1 and q2 are the charges, and r is the distance between the two charges.

## How do I determine the direction of the force on each charge?

The direction of the force on each charge can be determined by using the law of cosines to find the angle between the two charges and the distance r. The force will be attractive if the angle is less than 90 degrees and repulsive if the angle is greater than 90 degrees.

## What is the significance of the distances between the charges in the triangle?

The distances between the charges in the triangle determine the strength of the force between them. The closer the charges are to each other, the stronger the force will be.

## Can the force on one charge affect the force on the other charges in the triangle?

Yes, the force on one charge can affect the force on the other charges in the triangle. This is because the force between two charges is directly proportional to the product of the charges and inversely proportional to the distance between them. So, if one charge is moved closer or farther away, it will affect the force on the other charges.

## Are there any real-life applications of calculating force on charges in a triangle?

Yes, there are many real-life applications of calculating force on charges in a triangle. This concept is used in fields such as physics, engineering, and electronics to understand and predict the behavior of electrically charged particles in different situations, such as in circuits, electric motors, and particle accelerators.

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