Three point charges have equal magnitudes, two being positive and one negative.
ID: 2125085 • Letter: T
Question
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 6.0 ?C, and the lengths of the sides of the triangle are 1.5 cm. Calculate the magnitude of the net force that each charge experiences.
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 6.0 ?C, and the lengths of the sides of the triangle are 1.5 cm. Calculate the magnitude of the net force that each charge experiences.Explanation / Answer
For all the charges draw a free body diagram of each charge, drawing the force vectors in each direction so you know which direction they are pointing. Technically it doesnt matter whether you draw an arrow pointing toward a charge for a positive-positive interaction, or away from, as long as you use the same convention for each. (i.e. positive-positive, arrow points towards charge, then negative negative, force points away from charge.)
The Force on a charge is given by
F = kQq/r^2 where Q is charge one and q is charge 2.
lets examine the charge on the negative first and then by comparison we will know the other 2. Maybe....
the distance from the right charge to the bottom left is 4.6 cm = r
F = k (7.0)^2/r^2 = ? in the positive x direction. ( i dont wanna look up the value of k but it is coulombs constant.
The second force seems kinda tricky but by trig analysis we dont need to even type equations into a calculator
1) The Force pulls the charge up and to the right. by pathogorean we can get
4.6^2 = 2.3^2 + height of triangle
calculate height of triangle. This is the magnitude of the Force in the y direction once you divide it by the hypoteneuse
Fy is proportional to the height of the triangle/ 4.6 cm whcih = sqrt(3)/2
Fx = 1/2
now take the force you calculated for the x direction and multiply it by 1/2
do the same for the force you calculated in the x direction and multiply it by sqrt(3)/2
You this is the magnitude in each direction.
by pathagorean again
Fx^2 + Fy^2 = Ftot^2
= (Fx + 1/2Fx)^2 + (sqrt(3)/2Fy)^2 = Ftot^2
this is the force on the negative charge
To do the positive charge on the right it is the same thing except now there is still a Force in the negative F direction from the negative charge
and there is the same fraction of Forces from the top positive charge except The Fy force is down and the Fx force is to the right.
so (-Fx + 1/2Fx)^2 + (-Fy)^2 = Ftot^2 for the magnitude of the Force on the positive right charge.
For the top charge, use the same trigonometry as we used before but now the magnitude from the other charges are both in the x and y direction
1.) From the negative charge the Force is down and to the left
2.) From the negative charge the Force is up and to the right.
Since its an equilateral triangle, and the positive charge is being acted on by both a negative and a positive charge, The sum of the Forces will cancel and The Magnitude of the force on this charge is zero.
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