Unizor - Physics4Teens - Electromagnetism - Electric Field - Problems 1

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

Problems 1

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

Solution
F_P = −k·q²/d² (from q_P on q_R)
F_Q = −k·q²/d² (from q_Q on q_R)
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
F_P+Q = −k·q²·√3/d²

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
F_e is an electrical repelling force between objects.
T is a thread tension.
φ is an angle that each thread makes with a vertical.
F_e = k·q²/d² (Coulomb's Law)
T·cos(φ) = m·g (= weight)
T·sin(φ) = F_e (= repelling)
tan(φ) = F_e /(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
F_e is an electrical attracting force of proton to electron that keeps electron on its orbit.
F_e = k·q²/R² (Coulomb's Law)
ω is angular speed of an electron.
m·R·ω² = F_e (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)
ω ≅ 1.274·10¹⁷ rad/sec
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