course Phy 232
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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.
Student Answer: I am in University Physics
`q`qThe 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.
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.
Student Answer: I am in University Physics
`q`q65.0 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.
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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?
Student Answer: I am in University Physics
`q`qSTUDENT SOLUTION: For a part, to determine the electric potential a distance fo 2.5810^-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.
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Question: `qquery univ phy 23.58 (24.58 10th edition). Geiger counter: long central wire 145 microns radius, hollow cylinder radius 1.8 cm.
What potential difference between the wire in the cylinder will produce an electric field of 20,000 volts/m at 1.2 cm from the wire?
Student Answer:
E=20,000 volts/m
r=1.2 cm=.012m
b=1.8 cm
a=145 microns=.0145 cm
V=E*r*ln(b/a)
V=20,000 volts/m*.012m*ln(1.8cm /.0145cm)
V=1157.13 volts
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Given solution:
`q`q** The voltage V_ab is obtained by integrating the electric field from the radius of the central wire to the outer radius.
From this we determine that E = Vab / ln(b/a) * 1/r, where a is the inner radius and b the outer radius.
If E = 20,000 V/m at r = 1.2 cm then
Vab = E * r * ln(b/a) = 20,000 V/m * ln(1.8 cm / .0145 cm) * .012 m = 1157 V. **
Self-critique (if necessary): Ok
Self-critique Rating: 3
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Question: `qQuery univ 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?
Student Answer:
Radius of drop=.65 mm=.00065 m from r=infinity
Q=-1.2 pC=-1.2E-12 C
k=9*10^9 Nm^2/C^2*-1.2E-12 C/r^2=-.0108 Nm^2/C/r^2
E = kQ /r^2=9*10^9 Nm^2/C^2*-1.2E-12 C/r^2=-.0108 Nm^2/C/r^2
Integrate from r=.00065 to r=infinity (done on calculator)
E=-16.615 Nm/C
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Given Solution:
STUDENT RESPONSE FOLLOWED BY SOLUTION: The problem said that V was 0 at d = inifinity, which I understand 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 aligned along the surface of the raindrop. Probably all wrong. I tried several different integrand configurations 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.. **
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Self-critique (if necessary): I got it correct, I keep forgetting that a Nm/C is equal to a Volt. That makes a sense about the second drop of rain merging and radius=.00065*2^(1/3)
Self-critique Rating: 3
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&#Very good responses. Let me know if you have questions. &#