Query 27

#$&*

course Phy 202

4/25 7pm

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.

027.

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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.

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Your Solution:

-55V-135V = -190V

W = q*Vba = 190J/C*1.602x10^-19C = 3.0438x10^-17 J

180V/1e = 180eV

confidence rating #$&*:

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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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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qQuery Principles and General Physics 17.8: Potential difference required to give He nucleus

65.0 keV of KE.

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Your Solution:

65000eV = V*2e, therefore V*2e = 1.04x10^-14J, then V = 32500V

confidence rating #$&*:

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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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Self-critique (if necessary):

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Self-critique Rating:

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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?

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Your Solution:

V= [8.99x10^9Nm^2/C2 * 1.602x10^-19C]/2.5x10^-15m = 5.8x10^5V

for 2 protons

V= k*Q1*Q2/r = 9.2x10^-14V

confidence rating #$&*:

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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

The given solution can benefit from an expanded explanation:

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 is V = k Q / r, and is defined to be

the work per unit charge necessary to move another charge from infinite separation to that point.

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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Self-critique (if necessary):

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Self-critique Rating:

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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?

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Your Solution:

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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).