To solve this problem we will apply the concepts related to electric potential and electric potential energy. By definition we know that the electric potential is determined under the function:
= Coulomb's constant
q = Charge
r = Radius
At the same time
The values of variables are the same, then if we replace in a single equation we have this expression,
If we replace the values, we have finally that the charge is,
Therefore the potential energy of the system is
Answer:
4.9x10^-6T
Explanation:
See attached file
Answer:
P /K = 1,997 10⁻³⁶ s⁻¹
Explanation:
For this exercise let's start by finding the radiation emitted from the accelerator
=
the radius of the orbit is the radius of the accelerator a = r = 0.530 m
let's calculate
\frac{dE}{dt} = [(1.6 10⁻¹⁹)² 0.530²] / [6π 8.85 10⁻¹² (3 108)³]
P= \frac{dE}{dt}= 1.597 10⁻⁵⁴ W
Now let's reduce the kinetic energy to SI units
K = 5.0 10⁶ eV (1.6 10⁻¹⁹ J / 1 eV) = 8.0 10⁻¹⁹ J
the fraction of energy emitted is
P / K = 1.597 10⁻⁵⁴ / 8.0 10⁻¹⁹
P /K = 1,997 10⁻³⁶ s⁻¹