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

__Problems 1__

*Problem A*

At three vertices

**,**

*P***and**

*Q***of an equilateral triangle with the length of a side**

*R**there are three electrical charges*

**d****,**

*q*_{P}**and**

*q*_{Q}**. All charges are equal in magnitude to**

*q*_{R}*, but the first two are positive, while the third is negative:*

**q**

*q*_{P}= q_{Q}= −q_{R}= qDetermine the magnitude and direction of the combined electrical attraction force acting on charge

*from*

**q**_{R}*and*

**q**_{P}*.*

**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

*to a midpoint between*

**R***and*

**P***.*

**Q**Its magnitude is

**F**_{P+Q}= −k·q²·√3/d²*Problem B*

Two point-objects of mass

*each are hanging at the same level above the ground on two parallel vertical threads in the gravitational field with a free fall acceleration*

**m***.*

**g**When they are charged with the same amount of electricity

*, they move away from each other to a distance*

**q***from each other, and each thread will make some angle with a vertical.*

**d**Determine this angle as a function of known parameters.

*Solution*

*is an electrical repelling force between objects.*

**F**_{e}*is a thread tension.*

**T***is an angle that each thread makes with a vertical.*

**φ***(*

**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

*, for an electron it is*

**+q***.*

**−q**The mass of an electron is

*.*

**m**What is the angular speed and frequency of rotation of electron?

*Solution*

*is an electrical attracting force of proton to electron that keeps electron on its orbit.*

**F**_{e}*(*

**F**_{e}= k·q²/R²*Coulomb's Law*)

*is angular speed of an electron.*

**ω***(*

**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 = 9.0·10**N·m²/C²^{9}

**q = 1.602·10**C^{−19}

**m = 9.109·10**kg^{−31}*(empirical)*

**R = 25·10**m^{−12}

**ω ≅ 1.274·10**rad/sec^{17}

**ν ≅ 2.028·10**rev/sec^{16}*Problem D*

A proton of mass

**m=1.673·10**kg^{−27}

**q=1.602·10**C^{−19}

**a=0.5**m*protons.*

**N=92**What should be the minimal initial speed

*of bombarding proton to overcome the repelling force of all protons inside the nucleus of Uranium and get closer to it on a distance*

**v***from a nucleus?*

**b=0.001**m*Solution*

Initial kinetic energy of a bombarding proton

*should be equal to a work*

**E=m·v²/2***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.*

**A**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

*be a variable distance of a proton to a nucleus.*

**x**

**F(x) = k·q·N·q/x²=k·N·q²/x²**Then the work done by this force on a segment from

*to*

**x***is*

**x+**d**x***d*

**A = k·N·q²·**d**x/x²**The total amount of work done by electrical repelling force is

∫

_{[a,b]}

*[*

**k·N·q²·**d**x/x² =**

= k·N·q²·= 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
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