# Magnitude of net force

Three positive particles of charges Q = 61.8 micro coulombs are located at the corners of an equilateral triangle with a side L = 16.3cm. Calculate the magnitude of the net force on each particle.

## Solution This solution is **FREE** courtesy of BrainMass!

Use Coulomb's Law to get the force exerted by one charge on the other:

F = k * q1 * q2 / d^2

F is the magnitude of the force. The direction of the force on what particle is in the opposite direction of the other particle when the charges are of the same sign.

k is Coulomb's constant--9 * 10^9 Newton meters^2 /

Coulomb^2

q1 and q2 are equal and so are 61.8 * 10^-6 Coulombs.

d is the distance between charges 16.3 * 10^-2 meters.

So between a single pair of charges the force is

F =

(9 * 10^9 N*m^2 / C^2)*(61.8^2 * 10^-12 C^2) / (16.3^2

* 10^-4 m^2) =

1.29*10^3 N

Now, for a particular reference particle, imagine its

location is the origin. Imagine the two other

particles are d meteres away, one of which is 30

degrees clockwise from the x-axis, and the other is 30

degrees counterclockwise from the x axis.

Then the force vector from the first particle is

pointing at 30 degrees clockwise from the negative x

axis. then, the force vector from the second particle

is pointing at 30 degrees counterclockwise from the

negative y axis.

When you add these two vectors, the y-axis components

sum to zero. The projection of each of the vectors to

the negative x axis is F times the cosine of 30

degrees, or about F * .866.

So, the total of the two force vectors on the reference

particle is 2 * F * .866 in the negative x-direction

and 0 in the y-direction. The magnitude of the

total force vector is simply

F = 2.0 * 1.29 * 10^3 * .866 N

= 2.23 * 10^3 N.