Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed...

Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equilateral triangle, as the drawing shows. The magnitude of each of the charges is 3.5 μC, and the lengths of the sides of the triangle are 2.0 cm. Calculate the magnitude of the net force that each charge experiences.

Expert Solution

The force experienced by the positive charge (say at the top corner of the equilateral triangle )due to another positive charge is
F = k q² / r²
Where q is the charge and r is the distance between the charges
F = 9 x 10⁹ x (3.5 x 10^-6)² / (0.02 m)²
F =  275.62 N
This is directed away from the positive charge along the line joining the two charges
The force on the positive charge due to the negative charge is
F =  k q² / r²
F =  9 x 10⁹ x (3.5 x 10^-6)² / (0.02 m)²
F =  275.62 N
This force directed towards the negative charge
To find the net force on the positive charge
When we make the forces in their component form, we can see that the vertical components (F sin ) cancels and the parallel components gets added
The net force is
F_net = 2 F cos
Where is the angle between the force and the horizontal direction.
F_net = 2 x  275.62 x cos 60⁰
F_net =   275.62 N
Directed horizontally towards the right or left (depending on where the negative charge resides)
The force experienced by the other positve charge due to positive and negative charges is the same as the above case.
For the negative charge, the force is due to both positive charges
The force experienced is
F = F =  k q² / r²
F =  9 x 10⁹ x (3.5 x 10^-6)² / (0.02 m)²
F =  275.62 N
Both the forces are directed towards the positve charge.
The two forces are separated by an angle of 60⁰. Thus the net force is
F_net = sqrt (F² + F² + 2 F² cos 60⁰)
F_net = sqrt(3) F
F_net = 1.732 x 275.62 N = 477.39 N
F_net = 477.39 N

genius_generous answered 2 years ago

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