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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.