# Calculate Speed of Electron from Charge Density

• tony873004
In summary, the speed of the electron can be calculated using v = √(2kqQ/mr) with q, Q, and m as the charge and mass values given in the problem.
tony873004
Gold Member
** 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?

Last edited:
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.

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.

## 1. How is the speed of an electron calculated from charge density?

The speed of an electron can be calculated by using the formula v = √(2qE/m), where q is the charge of the electron, E is the electric field, and m is the mass of the electron.

## 2. What is charge density and how does it affect the speed of an electron?

Charge density is the amount of electric charge per unit volume. It can affect the speed of an electron by determining the strength of the electric field that the electron is moving through.

## 3. Can the speed of an electron be calculated using other methods besides charge density?

Yes, the speed of an electron can also be calculated by using the kinetic energy formula, KE=(1/2)mv^2, where m is the mass of the electron and v is the speed.

## 4. How does the speed of an electron compare to the speed of light?

The speed of an electron is much slower than the speed of light. The maximum speed an electron can reach is about 1/10,000th the speed of light.

## 5. What are some real-world applications of calculating the speed of an electron from charge density?

Understanding the speed of electrons is crucial in many fields such as electronics, telecommunications, and particle physics. It can also be used to determine the conditions and properties of materials and their conductivity.

Replies
2
Views
1K
Replies
2
Views
2K
Replies
10
Views
1K
Replies
11
Views
4K
Replies
1
Views
1K
Replies
2
Views
1K
Replies
10
Views
494
Replies
6
Views
995
Replies
2
Views
8K
Replies
9
Views
6K