Determining the electron speed in parallel plates

[ProAgnusDei]

Homework Statement

Two parallel plates labelled W (negative) and X (positive) are separated by 5.2 cm. The electric potential between the plates is 150 V. An electron starts from rest at time t_w and reaches plate X at time t_x. The electron continues through an opening in plate X and reaches point P at time t_P. Determine the speed at t_x and t_P

e = -1.6 x 10^-19 C
m_e = 9.1 x 10^-31

Homework Equations

V = kq₁ / r
ΔV = ΔE/q
ε = ΔV / r
E_k = 0.5mv²

The Attempt at a Solution

I'm not sure how to determine v at t_w and t_x, but I did find the field strength: ε = 2884.6. We also know the electron starts from rest, so we could use the conservation of energy law to generate the following eq'n.

v² = 2m^-1kq₁q₂(r_a^-2 - r_b²). But we don't have a "q₂" so I'm quite lost. Any help is appreciated!

God bless and thanks!

 

[gneill]

The Volt is a name given to the unit combination Joules/Coulomb

You should be able to find (in your notes or text) an expression for the work done on a charge falling through a potential difference. The work done shows up as kinetic energy...

 

[NewtonsHead]

You can start by finding the acceleration of the electron.
F = ma = qE
a = qE/m
Then it's a kinematics problem
vf^2 = 2*a*x Thats good for finding Vx

 

[ProAgnusDei]

I used your equation (not on my eq'n sheet), a = qE/m, substituted Vq for E, leaving me with Vq˄2 / m. I found acc. to be 3.376 x 10^-6.

Plugged that into eq'n v₂² = v₁² + 2ad,

which reduces to v₂ = sqr(2ad), and I got a velocity (v_x) of 5.93 x 10^-4 m/s which seems really slow to me.

 

[gneill]

I used your equation (not on my eq'n sheet), a = qE/m, substituted Vq for E, leaving me with Vq˄2 / m. I found acc. to be 3.376 x 10^-6.

Why would you substitute Vq for E? E here would be the electric field strength (units of [N]/[C] or [V]/[m]). You calculated the electric field strength previously, calling it ε.

However, a more slick approach is to use q*ΔV, with q being the charge on the electron and ΔV the change in electric potential (the 150V). The units are then [C][J]/[C] → [J], which is energy. That's the work done on the electron as it falls through the field...

 

[ProAgnusDei]

Ok that makes more sense. When you wrote E, I figured you meant energy, not ε. Tak.