Two identical charged particles of mass m=1.6 kg and electric charge Q=-5.2 μC are placed on an frictionless 26° incline. One of them falls down to the bottom and stops against the wall. The other stays at rest in the middle of the incline as shown in Figure. Find is the distance between the particles.

The charged particle on the incline stays in the middle because the electrstatic force of repulsion balances the gravitational force acting along the incline downwards.

If the distance between the charged particles is x:

k Q^2/x^2 = mg*(sin21deg)

here k =1/4*pi*epsilon0
Compute x from the above expression.

To find the distance between the particles, we first need to analyze the forces acting on each particle and calculate their respective accelerations.

Let's start with the particle that falls down to the bottom. Since it stops against the wall, it means the net force acting on it is zero. The forces acting on this particle are the gravitational force (mg) acting vertically downwards and the electrostatic force (Fe) acting vertically upwards due to the repulsion between the charged particles.

The force of gravity can be resolved into components along and perpendicular to the incline. The component along the incline is mg*sin(θ), where θ is the angle of the incline (26° in this case), and the component perpendicular to the incline is mg*cos(θ). Since the incline is frictionless, there is no force parallel to the incline.

The electrostatic force between the two charged particles can be calculated using Coulomb's Law: Fe = k * (Q1 * Q2) / r^2, where k is the Coulomb's constant (8.99 × 10^9 N*m^2/C^2), Q1 and Q2 are the charges of the particles (-5.2 μC in this case), and r is the distance between the particles.

Now, let's calculate the acceleration of the falling particle. Since the net force is zero, the electrostatic force (Fe) must balance the component of the gravitational force along the incline (mg*sin(θ)). Therefore, we have:

Fe = mg*sin(θ)
k * (Q^2) / r^2 = mg * sin(θ)

Now, let's move on to the particle that stays at rest in the middle of the incline. The forces acting on this particle are the gravitational force (mg) acting vertically downwards and the electrostatic force (F') acting vertically upwards due to the repulsion from the falling particle. Similar to the previous case, we can resolve the gravitational force into components along and perpendicular to the incline. Again, since the incline is frictionless, there is no force parallel to the incline.

The electrostatic force between the two particles in this case can also be calculated using Coulomb's Law: F' = k * (Q^2) / (2r)^2, where 2r represents the distance between the particles (since the falling particle is at the bottom, and the other particle is in the middle, the distance between them is twice the distance between the resting particle and the bottom).

Now, let's calculate the acceleration of the resting particle. The net force acting on it is the electrostatic force (F') minus the component of the gravitational force along the incline (mg*sin(θ)). Therefore, we have:

F' - mg*sin(θ) = ma'

where a' is the acceleration of the resting particle.

Since we want the resting particle to stay at rest, its acceleration (a') should be zero. Hence, we have:

F' - mg*sin(θ) = 0

Now we have two equations: one relating the electrostatic force (Fe) of the falling particle to the distance between the particles (r), and another equation relating the electrostatic force (F') of the resting particle to the same distance. By solving these equations simultaneously, we can find the distance (r) between the particles.

Let's substitute the expressions for Fe and F' into their respective equations:

k * (Q^2) / r^2 = mg * sin(θ)
k * (Q^2) / (2r)^2 - mg*sin(θ) = 0

Now, we can solve these equations for the distance (r) between the particles. By substituting the known values for mass, charge, angle, and Coulomb's constant, you can calculate the distance between the particles.