physics2 collage

Question

Two co-axial cylindirical conducting shells with radius a and b has lenghts L. If these cylindirical shells

have ±Q charges on them. a=2m, b=4m, Q=143C, L=5m and

a mass of 2kg that has a charge of 1C is released near

the inner conductor, what would its velocity be when

it reaches the outer conductor?

Answer 1

A positive point charge q of mass m is released from rest in a uniform electric field E and directed along the

x-axis. The acceleration of the charged mass is constant and given by qE/m. The motion is simple linear motion along the x-axis.

One can then apply the equations of kinematics in one dimension:

x − x₀ = v₀t + 0.5at² and v = d(v₀t + 0.5at²)/dt or v₀ + at.

Also, v² = v₀² + 2av₀t + a²t², rewritten as v₀² + 2a(x − x₀ − 0.5at²) + a²t² which reduces to v₀² + 2a(x − x₀).

Now take x₀=0 & v₀=0 to obtain (from x − x₀ = v₀t + 0.5at²) x = 0.5at² or (qE/2m)t². Next, v or dx/dt equals (qE/m)t.

Also, note from v² = v₀² + 2a(x − x₀) that v² = 2ax at x₀=0 & v₀=0. Since a or dv/dt = d(qE/m)t/dt or qE/m, it follows that v² = (2qE/m)x.

By way of Gauss' Law, the electric field between the cylindrical shells is obtained as λ/2πε₀r where λ=143 Coulombs/5 Meters, ε₀=8.854187817E-12 Coulomb/Newton•Meter², and r = (4 − 2) Meters. These values give E equal to 2.570439811E11 Newtons Per Coulomb.

Place E=2.570439811E11 Newtons Per Coulomb into v² = (2qE/m)x where q = 1 Coulomb, m = 2 Kilograms, and x = 2 Meters to yield Velocity equal to 716999.2763 Meters Per Second.