A thin cylindrical shell of radius R1 = 5.5 cm

Question

A thin cylindrical shell of radius R1 = 5.5 cm is surrounded by a second cylindrical shell of radius R2 = 9.0 cm, as in (Figure 1). Both cylinders are 15 mm long and the inner one carries a total charge Q1= -0.68 nC and the outer one Q2 = +1.56 nC.

(a) If an electron (m=9.1×10−31kg) escaped from the surface of the inner cylinder with negligible speed, what would be its speed when it reached the outer cylinder?

Express your answers with the appropriate units.

(b) If a proton (m=1.67×10−27kg) revolves in a circular orbit of radius R = 7.0 cm about the axis (i.e., between the cylinders), what must be its speed?

Answer 1

Hello Amalia!

I think it would be helpful to first draw a sketch! Be sure to define the two radii and which charges belong to which cylinders.

Useful definitions:

ΔU - Potential Energy Difference

ΔV = Voltage Potential Difference

V = Voltage

q = electric charge

m = mass

υ = particle velocity

Knowns:

R₁ = 5.5 [cm] = 5.5x10^-2 [m]

R₂ = 9 [cm] = 9x10^-2 [m]

L = 15 [mm]

Q₁ = -0.68 [nC] = -0.68x10^-9 [C]

Q₂ = +1.56 [nC] = +1.56x10^-9 [C]

q_e = -1.6x10^-19 [C]

m_e = 9.1x10^-31 [kg]

m_p = 1.67x10^-27 [kg]

k = 9x10⁹ [Nm² / C²]

Useful Formulas:

ΔU = q_eΔV

V = k(q/r)

ΔV = V₂ - V₁

υ = sqrt(2qΔV / m) = sqrt(2ΔU / m)

As we know, potential energy can be converted to kinetic energy. In this case, the electron jumps from one cylinder to the other. This means that it converts its potential energy to kinetic, gaining velocity. (The potential in this case is electric potential)

So, lets plug in for Part A:

ΔU = q_eΔV = q_e(V₂ - V₁) = q_ek [ (Q₂/r₂) - (Q₁/r₁) ]

= (1.6x10^-19 [C] )(9x10⁹ [Nm² / C²] ) * [ (+1.56x10^-9 [C] / 9x10^-2 [m]) - (-0.68x10^-9 [C] / 5.5x10^-2 [m]) ]

∴ ΔU ~ 4.28x10^-17 [J]

Now plug in to the velocity formula:

υ_e = sqrt(2ΔU / m_e) = sqrt( 2*(4.28x10^-17 [kgm²/s²] / 9.1x10^-31 [kg] )

∴υ_e ~ 9.7x10⁶ [m/s]

For Part B:

Recall that F = ma, but for circular orbits, we can state: F = m (v² / r)

In this case, F can be represented as the F_E, or the force due to the electric field.

Knowns:

ξ₀ = 8.85x10^-12 [F/m]

m_p = 1.67x10^-27 [kg]

q_p = 1.6x10^-19 [C]

r₃ = 7 [cm] = 7x10^-2 [m]

Formulas:

F = m (v² / r)

A = 2πLr

F_E = Q / ξ₀A

Solving for velocity, we get:

∴ υ = sqrt(q_pQ₁r₃ / 2πLr₁m_pξ₀)

Now just plug in! Hope this helps! If you have anymore questions, please feel free to contact me!