Open Query 27

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course Phy 122

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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:

dV= -190 *1.6e-19=3.04e-17J

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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): OK

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Self-critique Rating: OK

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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:

6.5e4 voles

Confidence rating: 3

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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.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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