## Friday, January 10, 2020

### Unizor - Physics4Teens - Electromagnetism - Electric Field - Problems 1

Notes to a video lecture on http://www.unizor.com

Problems 1

Problem A
At three vertices PQ and R of an equilateral triangle with the length of a side d there are three electrical charges qPqQ and qR. All charges are equal in magnitude to q, but the first two are positive, while the third is negative:
qP = qQ = −qR = q
Determine the magnitude and direction of the combined electrical attraction force acting on charge qR from qP and qQ.

Solution
FP = −k·q²/ (from qP on qR)
FQ = −k·q²/ (from qQ on qR)
These forces are at angle π/3 to each other. Adding them as vectors. The direction of a combined force is from point R to a midpoint between P and Q.
Its magnitude is
FP+Q = −k·q²·√3/

Problem B
Two point-objects of mass m each are hanging at the same level above the ground on two parallel vertical threads in the gravitational field with a free fall acceleration g.
When they are charged with the same amount of electricity q, they move away from each other to a distance d from each other, and each thread will make some angle with a vertical.
Determine this angle as a function of known parameters.

Solution
Fe is an electrical repelling force between objects.
φ is an angle that each thread makes with a vertical.
Fe = k·q²/d² (Coulomb's Law)
T·cos(φ) = m·g (= weight)
T·sin(φ) = Fe (= repelling)
tan(φ) = Fe /(m·g)
tan(φ) = k·q² /(m·g·d²)

Problem C
We assume the Bohr's classical model of an atom.
The atom of hydrogen has 1 proton and 1 electron rotating around it on a distance R.
The electric charge of a proton is positive +q, for an electron it is −q.
The mass of an electron is m.
What is the angular speed and frequency of rotation of electron?

Solution
Fe is an electrical attracting force of proton to electron that keeps electron on its orbit.
Fe = k·q²/R² (Coulomb's Law)
ω is angular speed of an electron.
m·R·ω² = Fe (Newton's Law)
ω = √k·q²/(R³·m)
ν = ω/(2π)
Let's substitute real data (all numbers are in SI units):
k = 9.0·109 N·m²/C²
q = 1.602·10−19 C
m = 9.109·10−31 kg
R = 25·10−12 m (empirical)
ν ≅ 2.028·1016 rev/sec

Problem D
A proton of mass m=1.673·10−27kg and charge of q=1.602·10−19C is launched from a distance a=0.5m towards a nucleus of Uranium that has N=92 protons.
What should be the minimal initial speed v of bombarding proton to overcome the repelling force of all protons inside the nucleus of Uranium and get closer to it on a distance b=0.001m from a nucleus?

Solution
Initial kinetic energy of a bombarding proton E=m·v²/2 should be equal to a work A of electric repelling force between a bombarding proton and the protons inside a nucleus. This work is done by a variable repelling force that depends on a distance between bombarding proton and a nucleus.
Since the force depends on the distance, we have to integrate the work done by this force along the distance covered by a bombarding proton.
Let x be a variable distance of a proton to a nucleus.
F(x) = k·q·N·q/x²=k·N·q²/x²
Then the work done by this force on a segment from x to x+dx is
dA = k·N·q²·dx/x²
The total amount of work done by electrical repelling force is
[a,b]k·N·q²·dx/x² =
= k·N·q²·
[(1/b)−(1/a)]
This is supposed to be equal to kinetic energy of a bombarding proton at the beginning of its motion
m·v²/2 = k·N·q²·[(1/b)−(1/a)]
This gives the value of speed v
v = √2·k·N·q²·[(1/b)−(1/a)]/m
Substituting the values listed above:
v ≅ 68.64 m/sec