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Wednesday, September 30, 2015

25: Problem 25.59

no title provided
INTRO:
Two 3.40g point charges on 1.00-m-long threads repel each other after being equally charged, as shown in the figure.Assume that θ = 28

Part A: 
What is the charge q?
Express your answer to three significant figures and include the appropriate units.

SOLUTION:
q = -1.32 μC

25: Problem 25.57

no title provided
INTRO:
You sometimes create a spark when you touch a doorknob after shuffling your feet on a carpet. Why? The air always has a few free electrons that have been kicked out of atoms by cosmic rays. If an electric field is present, a free electron is accelerated until it collides with an air molecule. It will transfer its kinetic energy to the molecule, then accelerate, then collide, then accelerate, collide, and so on. If the electron’s kinetic energy just before a collision is 1.80×10−18J or more, it has sufficient energy to kick an electron out of the molecule it hits. Where there was one free electron, now there are two! Each of these can then accelerate, hit a molecule, and kick out another electron. Then there will be four free electrons. In other words, as shows below, a sufficiently strong electric field causes a “chain reaction” of electron production. This is called a breakdown of the air. The current of moving electrons is what gives you the shock, and a spark is generated when the electrons recombine with the positive ions and give off excess energy as a burst of light.

Part A: 
The average distance an electron travels between collisions is 2.00 μm . What acceleration must an electron have to gain 1.80×10−18J of kinetic energy in this distance?
Express your answer to three significant figures and include the appropriate units.

SOLUTION:
a = 9.88×1017 m/s2

Part B:
What force must act on an electron to give it the acceleration found in part A?
Express your answer to three significant figures and include the appropriate units.

SOLUTION:
F = 9.00×10-13 N

Part C:
What strength electric field will exert this much force on an electron? This is the breakdown field strength. 
Note: The measured breakdown field strength is a little less than your calculated value because our model of the process is a bit too simple. Even so, your calculated value is close.
Express your answer to three significant figures and include the appropriate units.

SOLUTION:
E = 5.63×106 N/C

Part D:
Suppose a free electron in air is 0.700 cm away from a point charge. What minimum charge must this point charge have to cause a breakdown of the air and create a spark? Assume the electron is not hitting the charge particle but a neutral air atom that is still 2.00 μm away.
Express your answer to three significant figures and include the appropriate units.

SOLUTION:
q = 30.6 nC

25: Problem 25.51

INTRO:
In a simple model of the hydrogen atom, the electron moves in a circular orbit of radius 0.053 nm around a stationary proton.

Part A: 
How many revolutions per second does the electron make?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
6.6×1015 rev/s

UPDATE, 09/12/2016

The first thing I always do is convert my givens to SI units 
r = 0.053 nm = 5.3×10-2 nm = 5.3×10-11 m

Now. We know an ELECTRON is moving around a PROTON, so we can look up the charges for these two items
q1 = qPROTON = 1.602×10-19 C
q2 = qELECTRON = -1.602×10-19 C

Next we need to do a basic dynamics problem. 
There is an electromagnetic force pulling the electron around the neutron. F = ma
the acceleration is tangential to the orbit, so a = an = ω2⋅r
ω = v/r ∴ ω2 = v2 / r2 ∴ an = v2 / r2 ⋅ r = v2 / r
⇒ F = m⋅an = m⋅v2 / r

Since we're trying to find the velocity and know everything else except the force (and this entire section has to do with Force & Coulomb's Law...)
F = ke⋅q1⋅q2/r2
∴ ke⋅q1⋅q2/r2 = m⋅v2 / r

⇒ ke⋅q1⋅q2 / (r⋅m) = v2
∴ v = (ke⋅q1⋅q2 / (r⋅m) )½

​
The mass of an electron is me = 9.109×10-31 kg
v = (ke⋅q1⋅q2 / (r⋅m) )½ 
= SQRT{ (8.99×109 Nm2/C2) ⋅ (1.602×10-19 C) ⋅ (-1.602×10-19 C) ⋅ 1/(9.109×10-31 kg) ⋅ 1/(5.3×10-11 m) ⋅ (kg⋅m/s2)/N }
= SQRT{ -4.779 × 1012 m2 / s2 }
--> v = 2.186 × 106 m/s

convert to rev/s -> 1 rev / (2 π r) = 1 rev / 3.33×10-10 m

v = 6.565 rev/s
*update (07/26/2017 @ Kirtan Jani) * v = 6.565×1015 rev/s =  6.6×1015 rev/s
This was wrong because I missed an exponent while explaining. However, if you look at the purple highlighted answer above.... you can see that that was only wrong because you were being too GD judgemental.  Remember this is a free service bruh.. 

25: Problem 25.40

no title provided
Part A: 
What is the force F⃗ on the 1 nC charge at the bottom?
Write your answer as two vector components, separated by a comma. Express each component numerically, in newtons, to two significant figures.

SOLUTION:
Fx, Fy = 0, 1.1×10-5 N

25: Problem 25.33

no title provided
Part A: 
What is the magnitude of the force F⃗ on the 1.0 nC charge in the figure?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
F = 3.1×10-4 N

Here's why ->


Part B:
What is the direction of the force F⃗ on the 1.0 nC charge in the figure?

SOLUTION:
Upward



25: Problem 25.9

INTRO:
Two neutral metal spheres on wood stands are touching. A negatively charged rod is held directly above the top of the left sphere, not quite touching it. While the rod is there, the right sphere is moved so that the spheres no longer touch. Then the rod is withdrawn.

Part A: 
Afterward, what is the charge state of each sphere?

a) Both the spheres are neutral.
b) The left sphere is negatively charged, the right sphere is charged positively.
c) The right sphere is negatively charged, the left sphere is charged positively.
d) Both the spheres are charged positively.
e) Both the spheres are charged negatively.

SOLUTION:
option c)

25: Problem 25.30

INTRO:
The nucleus of a 125Xe atom (an isotope of the element xenon with mass 125 u) is 6.0 fm in diameter. It has 54 protons and charge q = +54e.

Part A: 
What is the electric force on a proton 1.4 fm from the surface of the nucleus? 
Hint: Treat the spherical nucleus as a point charge.
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
We know the formula for the force between two charged particles is F = K |q1| |q2| r-2
We simply have to determine which values to plug into this equation and we will get the answer that we're looking for. 
Begin by taking the radius of the nucleus, which is half the diameter, or 3 fm
the added distance of the proton is 1.4 fm, so the total distance of the proton from the center of the nucleus is 3 + 1.4 = 4.4 fm = 4.4⋅10-15 m
so r = 4.4⋅10-15 m​
Next we need to determine the two charges. 
We're given that the nucleus has a charge of q = +54e. 
Knowing that e=1.6⋅10-19 C , qnucleus = 8.64⋅10-18 C
q1 = 8.64⋅10-18 C​
lastly, we know that the charge of ONE proton is just e, so
q2 = 1.6⋅10-19 C​
Now we have all of our values to plug in and can solve for the force
(remember K is the electrostatic constant and equal to 9.0⋅109 Nm2/C2)
∴F = (9.0⋅109 Nm2/C2)(8.64⋅10-18 C)(1.6⋅10-19 C)(4.4⋅10-15 m)-2​
→F = 642.645 N​

F = 643 N

Part B:
What is the proton's acceleration?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
We know that F=ma, so if we divide the force by the given mass...  a = 3.8×1029 m/s2