a point charge -q, mass m is flying up at the speed v(0)= v 0 from the distance x(0)=L from a uniformly and positively charged thin plate, its linear density equal p. determine the maximum distance of a particle from the plate
To determine the maximum distance of the particle from the plate, we need to consider the dynamics of the particle's motion under the influence of the electric force from the plate.
Let's break down the solution into steps:
Step 1: Determine the electric field due to the positively charged plate.
We can use Gauss's Law to find the electric field due to a uniformly charged thin plate. Since the plate has a linear charge density p, the electric field near the plate is given by:
E = σ/2ε₀
where E is the electric field, σ is the charge density, and ε₀ is the permittivity of free space.
Step 2: Analyze the forces acting on the charged particle.
The electric force experienced by the charged particle is given by:
F = qE
where F is the force, q is the charge of the particle, and E is the electric field. Since the force is in the opposite direction to the motion of the particle (upwards), the negative sign is included.
Step 3: Solve the equations of motion for the particle.
We can use Newton's second law to describe the motion of the particle. The force F (electric force) is the only force acting on the particle, so we'll have:
F = ma
where m is the mass of the particle and a is the acceleration.
Substituting the electric force equation into Newton's second law gives:
qE = ma
Step 4: Solve the differential equation for motion.
Assuming the initial velocity of the particle is v₀, we can rewrite the equation of motion as:
qσ/(2ε₀) = m(dv/dt)
where v is the velocity of the particle.
Integrating both sides of the equation gives:
∫(dv / v₀ to v) = (qσ / 2ε₀)∫(dt / 0 to t)
Solving the integral on both sides gives:
ln(v) - ln(v₀) = (qσ / 2ε₀)t
Simplifying further:
ln(v / v₀) = (qσ / 2ε₀)t
Step 5: Find the maximum distance.
To find the maximum distance, we need to find the time at which the velocity of the particle becomes zero. This corresponds to the point where the particle reaches its maximum distance and starts to fall back towards the plate.
Setting v = 0 in our equation:
ln(0 / v₀) = (qσ / 2ε₀)t
Since the natural logarithm of 0 is undefined, we solve for t when v becomes very close to 0, which implies ln(v / v₀) approaches negative infinity.
This gives us:
(qσ / 2ε₀)t = -∞
From this equation, we can see that t = 0, as any finite value of t would give an undefined result.
Therefore, the maximum distance the particle reaches from the plate occurs at t = 0, which is its initial distance from the plate, L. Hence, the maximum distance is L.
Note: This solution assumes the particle starts from rest at x₀ = L. If the particle has an initial velocity, v₀, different from 0, the equations will be slightly modified, and the maximum distance will be different.