Assignment 27

course Phy 202

1:45am 7/25

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

Your Solution:

Change in potential= Potential final- potential initial

Change= -55 V- 135 V

Change= -190 V

The change in potential energy of the proton is,

(-190 V)(1.6*10^-19 C)= -3.0*10^-17 J

The work in electron volts=

180 volts*charge of 1 electron= 180 eV

confidence rating:

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2

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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

Self-critique Rating:

3

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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:

Energy= (65.0 *10^3 eV)= 6.50*10^4 eV

To gain 6.50 *10^4 eV of energy the voltage difference would therefore be ½(6.50*10^4 voles)= 3.35*10^4 volts

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 volts, or 3.35 * 10^4 volts.

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Self-critique (if necessary):

ok

Self-critique Rating:

2

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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= kq/r

V= (9.0*10^9 Nm^2)*(1.60*10^-19 C)/(2.5*10^-15 m)

V= 5.8*10^5 V

The potential energy is equal to the work done against the field

PE= (1.60*10^-19 C)(5.8*10^5 V)

PE= 9.2*10^-14 J

confidence rating:

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Given Solution: STUDENT 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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Self-critique (if necessary):

ok

Self-critique Rating:

3

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