MHB Answer: Calculate Electric Field Strength at 1mm for a Stationary Electron

  • Thread starter Thread starter Alexis Deleon
  • Start date Start date
  • Tags Tags
    Capacitance
Click For Summary
To calculate the electric field strength at a radial distance of 1 mm from a stationary electron, the charge q used is the negative elementary charge, approximately -1.602 x 10^-19 C. The electric field E can be determined using the formula E = kq/r^2, where k is the Coulomb's constant (1/(4πε₀)). The energy density U is then calculated using U = (1/2)(ε₀)(E^2). Substituting the values into the equations allows for the determination of U at the specified distance. This approach provides a clear method for calculating the electric field and energy density related to a stationary electron.
Alexis Deleon
Messages
1
Reaction score
0
Question: A stationary electron is a point of charge. What is the energy density of U, it's electric field at a radial distance r=1mm?

Know:
1mm=1*10^-3m
u=(1/2)(epsilon naught)(E^2)
E=kq/r^2

I know I have to end up solving for (E)^2, where E=kq/r^2, but how do I determine what charge to use for q?
 
Mathematics news on Phys.org
Alexis Deleon said:
Question: A stationary electron is a point of charge. What is the energy density of U, it's electric field at a radial distance r=1mm?

Know:
1mm=1*10^-3m
u=(1/2)(epsilon naught)(E^2)
E=kq/r^2

I know I have to end up solving for (E)^2, where E=kq/r^2, but how do I determine what charge to use for q?

Welcome to MHB Alexis Deleon! :)

The charge q is the negative elementary charge, usually denoted with the symbol $\text{e}$, which is also the first letter of the word electron:
$$q = -1 \text{ e} = -1.602 \cdot 10^{-19} C$$

Note also that:
$$k = \frac{1}{4\pi \epsilon_0}$$

So:
$$U=\frac 1 2 \epsilon_0 E^2
= \frac 1 2 \epsilon_0 \left( \frac {kq}{r^2}\right)^2
= \frac 1 2 \epsilon_0 \left( \frac{1}{4\pi \epsilon_0} \frac {-1\text{ e}}{r^2}\right)^2$$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

Similar threads

  • · Replies 4 ·
Replies
4
Views
961
  • · Replies 2 ·
Replies
2
Views
959
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
923
  • · Replies 7 ·
Replies
7
Views
2K