Particles 1 and 2 of charge q1 = q2 = +3.20 x 10-19 C are on a y axis at distance d = 22.0 cm from the origin. Particle 3 of charge q3 = +9.60 x 10-19 C is moved gradually along the x axis from x = 0 to x = +5.0 m.
At what value of x will the magnitude of the electrostatic force on the third particle from the other two particles be maximum?
What is the maximum magnitude?
Is particle 2 at x = 0, y = -22 cm?
If so, the net force on P3, F3, is always along the x axis, and is zero at x=0.
F3 = 2*k*q1*q3*x/(x^2 + a^2)^3/2
where a = 22 cm
Maximum force occurs where dF3/dx = 0
I get it to be where x = 22 cm/sqrt2 = 15.55 cm
See what you get, and use the equation above to get the maximum force F3
To find the value of x where the magnitude of the electrostatic force on the third particle is maximum, we need to calculate the net electrostatic force on the particle when it is at different positions along the x-axis and then find the position where the force is maximum.
The electrostatic force between two charged particles is given by Coulomb's law:
F = k * (q1 * q3) / r^2 (1)
where F is the electrostatic force, k is the Coulomb's constant (k = 8.99 x 10^9 N m^2/C^2), q1 and q3 are the charges of particles 1 and 3 respectively, and r is the distance between particles 1 and 3.
Now, let's calculate the force on particle 3 when it is at different positions along the x-axis:
When q3 is at x = 0:
The distance between particle 1 and 3 is r = sqrt(d^2) = sqrt(0.22^2) = 0.22 m.
Plugging these values into equation (1) gives:
F1 = (8.99 x 10^9) * (3.20 x 10^-19) * (9.60 x 10^-19) / (0.22^2) = 2.76 x 10^-5 N
When q3 is at x = +5.0 m:
The distance between particle 1 and 3 is r = sqrt((d^2) + (x^2)) = sqrt((0.22^2) + (5.0^2)) = 5.03 m.
Plugging these values into equation (1) gives:
F2 = (8.99 x 10^9) * (3.20 x 10^-19) * (9.60 x 10^-19) / (5.03^2) = 1.97 x 10^-6 N
Now, we can see that F2 < F1. Hence, the magnitude of the electrostatic force is maximum when q3 is at x = 0.
The maximum magnitude of the electrostatic force is F1 = 2.76 x 10^-5 N.
To find the value of x at which the magnitude of the electrostatic force on the third particle is maximum, we need to calculate the net force on the third particle due to the other two particles at different positions of particle 3 along the x-axis.
The electrostatic force between two charged particles can be calculated using Coulomb's Law:
F = k * |q1 * q2| / r^2
where F is the magnitude of the electrostatic force, k is the electrostatic constant (8.99 x 10^9 N*m^2/C^2), q1 and q2 are the charges of the particles, and r is the distance between the particles.
Let's start by calculating the force on particle 3 due to particles 1 and 2 at x = 0:
r1 = √((22.0 cm)^2 + (0 cm)^2) = 22.0 cm
F1 = k * |q1 * q3| / r1^2
Now, let's calculate the force on particle 3 due to particles 1 and 2 at x = +5.0 m:
r2 = √((22.0 cm)^2 + (5.0 m)^2) = √(484 cm^2 + 2500 cm^2) = √2984 cm ≈ 54.62 cm
F2 = k * |q1 * q3| / r2^2
We need to find the value of x where the magnitude of the electrostatic force on particle 3 is maximum. Therefore, we need to find the position where F2 - F1 is maximum.
To find this position, we compare the forces F1 and F2 and calculate the difference ΔF = F2 - F1 for different values of x along the x-axis, from 0 to 5.0 m. The position at which ΔF is maximum will give us the value of x where the magnitude of the electrostatic force on particle 3 is maximum.
Finally, the magnitude of the electrostatic force on particle 3 from the other two particles at this maximum position gives us the maximum magnitude of the force.
s particle 2 at x = 0, y = -22 cm?
If so, the net force on P3, F3, is always along the x axis, and is zero at x=0.
F3 = 2*k*q1*q3*x/(x^2 + a^2)^3/2
where a = 22 cm
Maximum force occurs where dF3/dx = 0
I get it to be where x = 22 cm/sqrt2 = 15.55 cm
See what you get, and use the equation above to get the maximum force F3