Assignment 6 Query

course PHY 122

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006.

Physics II

02-05-2007

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18:49:06

Query Principles and General Physics 17.4: work by field on proton from potential +135 V to potential -55 V.

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RESPONSE -->

The diiference of voltage potential between +135V and -55V is 190V.

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19:09:44

The change in potential is final potential - initial potential = -55 V - (125 V) = -180 V, so the change in the potential energy of the proton is {}{}-180 V * 1.6 * 10^-19 C ={}-180 J / C * 1.6 * 10^-19 C = -2.9 * 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 2.9 * 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 2.9 * 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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RESPONSE -->

I see what you were wanting now. A V = J/C so to find the work you would divide by the charge of the proton in coulombs That leaves you with J or work.

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19:21:25

Query Principles and General Physics 17.8: Potential difference required to give He nucleus 65.0 keV of KE.

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RESPONSE -->

I'm just not making any sense of this.

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19:32:59

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.

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RESPONSE -->

For some reason I can't see why The voltage difference would be half of the 6.5*10^4eV

I guess it's because I can't seem to come up with a definition of an eV. I thought I was doing fine until I started this query.

An electron volt is the energy required to move a charge e through an opposing 1 volt potential difference, where e is the magnitude of the charge on an electron.

This is also equal to the KE gained by a charge e as it 'fall's through a potential difference of 1 volt.

Here you have a helium nucleus, whose charge is 2 e, falling through a potential difference which gives it 65.0 keV of kinetic energy.

A charge of e falling through 65.0 keV would gain 65.0 keV of kinetic energy. If a charge of 2 e fell through this potential difference it would gain twice this much KE, or 130 keV.

A helium nucleus, with is charge of 2 e, would have to fall through only 32.5 keV to gain 65.0 keV of kinetic energy.

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See my note and see if it helps. Let me know if you have additional questions.