If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Question:
`qQuery Principles and General Physics 17.4: work by field on proton from potential +135 V
to potential -55 V.
Your Solution:
Confidence rating:
Given Solution:
The change in potential is final potential - initial
potential = -55 V - (135 V) = -190 V, so the change in the potential energy of
the proton is
-190 V * 1.6 * 10^-19 C =
-190 J / C * 1.6 * 10^-19 C = -3.0 * 10^-17 J.
In the absence of dissipative forces this is equal and
opposite to the change in the KE of the proton; i.e., the proton would gain
3.09 * 10^-17 J of kinetic energy.
Change in potential energy is equal and opposite to the
work done by the field on the charge, so the field does 3.0 * 10^-17 J of work
on the charge.
Since the charge of the proton is equal in magnitude to that
of an electron, he work in electron volts would be 180 volts * charge of 1
electron= 180 eV.
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Question:
`qQuery Principles and General Physics 17.8: Potential difference required to give He
nucleus 65.0 keV of KE.
Your Solution:
Confidence rating:
Given Solution: 65.0 keV is 65.0 * 10^3 eV, or 6.50 * 10^4 eV, of
energy.
The charge on a He nucleus is +2 e, where e is the charge on
an electron. So assuming no dissipative
forces, for every volt of potential difference, the He nuclues would gain 2 eV
of kinetic energy.
To gain 6.50 * 10^4 eV of energy the voltage difference
would therefore be half of 6.50 * 10^4 voles, or 3.35 * 10^4 volts.
STUDENT QUESTION
I didn’t convert the keV into eV. What do these units even
mean??
INSTRUCTOR RESPONSE
k means 'kilo'; so a keV is 10^3 eV.
An electron volt is the PE change of an electron as it moves through a PE change
of +1 volt.
A Joule is the PE change of a Coulomb of negative charge as it moves through a
PE change of +1 volt.
Since the charge of an electron has magnitude 1.6 * 10^-19 Coulomb, an electron
volt is 1.6 * 10^-19 Joules.
Within the electron shell of an atom, and in many other applications, we are
dealing with charges of in small whole-number multiples of 1.6 * 10^-19 Coulomb,
moving between points where potential changes are anywhere from a few millivolts
to a few volts. The associated energy changes are more easily thought of in
terms of electron volts (typical values might range from .001 eV to around 100
eV) than ergs or Joules (where the same range might be, for example, 10^-17
Joules to 10^-22 Joules). Numbers between .001 and 100 are easy to think about,
an relate easily to the energy of a single electron moving through a potential
difference of a single volt. Numbers like 10^-17 and 10^-22 are harder to
imagine.
STUDENT QUESTION
I understand the formula, but didn’t know where to find He?
May have missed it in the notes……anyway
INSTRUCTOR RESPONSE
It is assumed to be general knowledge that a helium nucleus contains 2 protons, each with a positive charge equal in magnitude to the electron charge. However not everyone remembers this, and without this knowledge it would be difficult to get the entire solution.
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Question:
`qQuery gen phy text
problem 17.18 potential 2.5 * 10^-15 m
from proton; PE of two protons at this separation in a nucleus.
What is the electrostatic potential at a distance of 2.5 *
10^-15 meters?
Your Solution:
Confidence rating:
Given Solution: STUDENT SOLUTION:
For a part, to determine the electric potential a distance of
2.58 * 10^-15m away from a proton, I simply used the equation V = k q / r for
electric potential for point charge:
q = 1.60*10^-19C=charge on proton
V = kq/r = 9.0*10^9N*m^2/C^2(1.60*10^-19C) / (2.5*10^-15m) =
5.8*10^5V.
Part B was the more difficult portion of the problem. You have to consider a system that consists
of two protons 2.5*10^-5m apart.
The work done against the electric field to assemble these
charges is W = qV. The potential energy
is equal to the work done against the field.
PE=(1.60*10^-19C)(5.8*10^5V)
= 9.2*10^-14 J.
STUDENT QUESTION
OK, got the first part.
I follow the second, but it doesn't make much sense. So you multiplied your
answer from the first by the charge again.
INSTRUCTOR RESPONSE
Th
Suppose you have two protons separated by a large, effectively infinite distance. They repel one another, so to move the protons closer to one another you would have to do positive work against the conservative electrostatic field. Just as when you lift an object against the conservative force of gravity, when you do work by 'pushing' an object against a conservative electrostatic field you increase its electrostatic PE. So you increase the PE of the system by 'pushing' one proton toward the other.
The electrostatic potential at distance r from a point charge
Q
In the present case, the point charge Q is a proton and the electrostatic potential at a distance of 2.5*10^-15m from the proton is +5.8*10^5 volts, or +5.8 * 10^5 Joules / Coulomb. What this means is that to move charge from a large distance to this position requires +5.8 * 10^5 Joules for every Coulomb of charge you move.
Now if the charge we are moving to that position is another proton, its charge is +1.6 * 10^-19 Coulombs. It therefore requires +5.8 Joules / Coulomb * 1.6 * 10^-19 Coulombs = 9.2 * 10^-14 Joules of work to get it there. This work is the potential energy of the system, relative to infinite separation.
In symbols, V = k Q / r, and to move a second charge q to that point therefore requires work `dW = V * q = k Q q / r, and this is the electrostatic potential energy of the two-proton system.
WORTH KNOWING BUT NOT NECESSARY AT THIS POINT:
Now the two protons exert a lot of force on one another, trying to push one another away. At close distances they are held together by another force, not an electrostatic force. This force is called the 'strong' force. If the electrostatic force breaks free of the 'strong' force (as two protons would do almost instantly), the two protons will repel each other and the electrostatic force will quickly accelerate them to speeds close to that of light. From an energy point of view, the positive PE of the system will be converted to KE.
Protons do manage to stay in this kind of close proximity in stable nuclei. Both protons and neutrons exert the 'strong' force on one another, if they are close enough. So for example if you get two protons and two neutrons into this sort of proximity, you get more 'strong' force and the electrostatic forces won't be able to break it. You end up with the very stable nucleus of Helium-4.
STUDENT QUESTION: A little confused about the V=k*Q/r,
where it looks like I can change the Q to q and substitute?
INSTRUCTOR RESPONSE
There is nothing special about Q and q. The charge can be
called Q or q or q1 or q2, or anything else.
The statement
'The electrostatic potential at distance r from a point charge Q is V = k Q / r'
uses Q for the charge. Had the statement been
'The electrostatic potential at distance r from a point charge q1 is V = k q1 /
r'
it would have been equally valid.
Either definition tells you that whatever the number of symbol you choose to use
for the charge, that symbol goes between k and the /.
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Question:
`qQuery univ 23.80 11th edition 23.78 (24.72 10th edition). Rain drop radius .65 mm charge -1.2 pC.
What is the potential at the surface of the rain drop?
Your Solution:
Confidence rating:
Given Solution:
STUDENT RESPONSE FOLLOWED BY SOLUTION: The problem said that V was 0 at d =
inifinity, which I understnad to mean that as we approach the raindrop from
infinity, the potential differencegrows from 0, to some amount at the surface
of the raindrop. Because water molecules
are more positive on one side that the other, they tend to align in a certain
direction. Since positive charges tend
to drift toward negative charge, I would think that the raindrop, with its
overall negative charge, has molecules arranged so that their more positive
sides are pointing toward the center and negative sides will be alighed along
the surface of the raindrop. Probably
all wrong. I tried several differnet
integrand configuraitons but never found one that gave me an answer in volts.
SOLUTION:
You will have charge Q = -1.2 * 10^-12 C on the surface of a
sphere of radius .00065 m.
The field is therefore E = k Q / r^2 = 9 * 10^9 N m^2 / C^2
* (-1.2 * 10^-12 C) / r^2 = -1.08 * 10^-2 N m^2 / C / r^2.
Integrating the field from infinity to .00065 m we get
(-1.08 * 10^-2 N m^2 / C) / (.00065 m) = -16.6 N m / C =
-16.6 V.
If two such drops
merge they form a sphere with twice the volume and hence 2^(1/3) times the
radius, and twice the charge.
The surface potential is proportional to charge and
inversely proportional to volume. So the
surface potential will be 2 / 2^(1/3) = 2^(2/3) times as great as before.
The surface potential is therefore 16.6 V * 2^(2/3) = -26.4 volts, approx.. **
STUDENT QUESTION:
I knew that my answer was off by some factor because the E
decreased from when it was just one raindrop. I didn’t get
that you would multiply it by two because the volume increased.
??? Can you explain why you would use the ratio of volume to radius increase in
order to get the new E???
INSTRUCTOR RESPONSE:
E depends on the total charge and the radius.
When the two drops merge, their charges combine. This gives you double the
charge compared to a single drop.
We don't care about the volume, we care about the radius. However we know what
happens to the volume: it doubles.
So we use what we know about the volume to determine what happens to the radius:
A sphere with twice the volume of another has 2^(1/3) times the radius.
We end up with double the charge on a sphere with 2^(1/3) times the radius.
Since the potential is proportional to the charge and inversely proportional to
the radius, the potential changes by factor 2^(2/3).
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