Three charges Q1
= -2μC (0,0)cm, Q2
= 4 μC (0,10)cm
and Q3
= -3 μC (15,0)cm acts on one another. Find the
magnitude of the total electric force and the angle
that the total electric force makes with the positive
x-axis.
Well, let me crunch the numbers and clown around with the calculations for you!
To find the total electric force, we need to calculate the individual forces between each pair of charges and then add them up.
Let's start with the force between Q1 and Q2. The distance between them is 10 cm = 0.1 m. We can use Coulomb's law to find the force:
F12 = k * |Q1| * |Q2| / r^2
where k is the electrostatic constant (8.99 x 10^9 Nm^2/C^2). Plugging in the values:
F12 = (8.99 x 10^9 Nm^2/C^2) * (2 x 10^-6 C) * (4 x 10^-6 C) / (0.1 m)^2
Solving this equation gives us F12 ≈ 719.2 N.
Next, let's calculate the force between Q1 and Q3. The distance between them is 15 cm = 0.15 m. Using Coulomb's law:
F13 = (8.99 x 10^9 Nm^2/C^2) * (2 x 10^-6 C) * (3 x 10^-6 C) / (0.15 m)^2
Solving this equation gives us F13 ≈ 239.7 N.
Finally, let's calculate the force between Q2 and Q3. The distance between them is 15 cm = 0.15 m. Using Coulomb's law again:
F23 = (8.99 x 10^9 Nm^2/C^2) * (4 x 10^-6 C) * (3 x 10^-6 C) / (0.15 m)^2
Solving this equation gives us F23 ≈ 479.4 N.
Now, to find the total electric force, we need to add up the individual forces:
F_total = √(F12^2 + F13^2 + F23^2)
Plugging in the values we calculated:
F_total ≈ √(719.2^2 + 239.7^2 + 479.4^2)
Solving this equation gives us F_total ≈ 875.9 N.
To find the angle that the total electric force makes with the positive x-axis, we can use trigonometry:
θ = arctan(F23 / F12)
Plugging in the values:
θ ≈ arctan(479.4 N / 719.2 N)
Solving this equation gives us θ ≈ 35.4°.
So, the magnitude of the total electric force is approximately 875.9 N, and the angle it makes with the positive x-axis is around 35.4 degrees. However, keep in mind that my calculations could be a little clownish, so it's always a good idea to double-check them!