電気力とベクトル / Electric Force and Vectors

On really really busy days, and I have no topic, I'll resort to solving homework questions and posting them. It's a two-for-one deal really; I keep my at least 1 post per day, and I help myself learn the concept. Oh, and I guess if someone finds this online it'll help them too. Because when considering Physics, everyone loses.

Consider three point charges, Q1, Q2, and Q3 as shown in the isosceles triangle below:

What is the magnitude and direction of the electric force on Q3?

First thing you should do is find the remaining values for your triangle. You are already given the lengths of each side, so by using that, you can find the angles inside the triangle, which will be used later.

Recall the general form of the Pythagorean Theorem: The Law of Cosines:

a² + b² – 2*a*b*cos(C) = c²

We can use this formula to determine the interior angles:

0.50² + 0.50² – 2(0.50)(0.50)cos(α) = 0.36²

(0.36² - 0.50² - 0.50²) / - (2(0.50)(0.50)) = cos(α)

cos(α) = 0.7408

cos^-1(0.7408) = 42.2

0.36² + 0.50² – 2(0.50)(0.36)cos(β) = 0.50²

(0.50² - 0.36² - 0.50²) / - (2(0.50)(0.36)) = cos(β)

cos(β) = 0.36

cos^-1(0.7408) = 68.9

So α = 42.2 and β = 0.36. Remember that since this is an isosceles triangle (Two sides have the same length), that means that each side shares the same angle, saving us one more application of the law of cosines.

Now we find the vector components of the forces acting on Q3. Draw a free body diagram for reference.

Use Coulomb's Law to calculate the magnitude of F31 and F32, then use trigonometry to break them down into the vertical and horizontal components (This is why we calculated the angles before).

Recall:

F_e = kq₁q₂ / r²

F31 = 9e9 * 10e-6 * 50e-6 / 0.5^2

F31 = 18 N

F31x = sin(42.2) * 18 = 12.1 N

F31y = cos(42.2) * 18 = 13.3 N

F32 = 9e9 * 10e-6 * -80e-6 / 0.36^2

F32 = 56 N

F32x = cos(68.9) * 56 = 20.2 N

F32y = sin(68.9) * 56 = 52.2 N

F3x = 12.1 + 20.2 = 32.3 N

F3y = 13.3 - 52.2 = -38.9 N

For F3y, you subtract instead of adding because F32 is on the negative vertical axis.

By Pythagorean Theorem:

F3 = √(F3x)²(F3y)²

F3 = √(32.3)²(38.9)²

F3 = 50.6 N

tan(θ) = 38.9 / 32.3 = 1.204

tan^-1(1.204) = 50.3^ο below the horizontal.

It is below the horizontal because Q2 exerts a greater attractive force than the repulsive force exerted by Q1, so it tends to go towards Q2.