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

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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.
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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?
 
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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:
$$U=\frac 1 2 \epsilon_0 E^2
= \frac 1 2 \epsilon_0 \left( \frac {kq}{r^2}\right)^2
= \frac 1 2 \epsilon_0 \left( \frac{1}{4\pi \epsilon_0} \frac {-1\text{ e}}{r^2}\right)^2$$
 
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