Calculate Speed of Electron from Charge Density

[tony873004]

** Edit: Nevermind. I figured it out using the Work Energy theorem.

An electron is released from rest 1.0 cim above a uniformly charged infinite plane with a charge density of 10^-9C/m². What is the speed of the electron when it hits the plane?

my attempt:
Potential energy when it is released= kinetic energy when it hits.

kqQ/r = 0.5 mv²

isolate v:

$$v = \sqrt {\frac{{2kqQ}}{{m \cdot r}}}$$

This would work if I was given 2 point charges, but how do I do this with a charge density and a point charge?

 

[Nate810]

Answer:The speed of the electron when it hits the plane can be calculated using the equation v = √(2kqQ/mr), where q is the charge of the electron, Q is the charge density of the plane, and m is the mass of the electron. In this case, q=1.6x10-19 C, Q=10-9 C/m2, and m=9.1x10-31 kg. Plugging in these values gives a speed of 1.9x106 m/s.

 

[Moetasim]

I would first clarify the units being used for charge density. Is it in coulombs per square meter (C/m2) or in coulombs per cubic meter (C/m3)? This is important because the calculation for speed will differ depending on the units used.

Assuming it is in C/m2, we can use the equation for electric field intensity to find the force acting on the electron. The electric field intensity (E) at a distance (r) from a charged plane with charge density (ρ) is given by E = ρ/2ε0, where ε0 is the permittivity of free space.

Using this equation, we can calculate the electric field intensity at a distance of 1.0 cm from the plane as follows:

E = (10-9 C/m2)/2(8.85x10-12 C2/Nm2) = 5.65x1011 N/C

Next, we can use the equation for force (F) to find the force acting on the electron:

F = Eq = (5.65x1011 N/C)(1.6x10-19 C) = 9.04x10-8 N

Finally, we can use the work-energy theorem to calculate the velocity of the electron. The work done by the electric force is equal to the change in kinetic energy of the electron, which can be expressed as:

W = Fd = \frac{1}{2}mv^2

Solving for v, we get:

v = \sqrt {\frac{2Fd}{m}}

Substituting the values we calculated, we get:

v = \sqrt {\frac{2(9.04x10-8 N)(1 cm)}{9.11x10-31 kg}} = 2.36x105 m/s

Therefore, the speed of the electron when it hits the plane is 2.36x105 m/s.