course phys202
1.work by field on proton from potential +135 V to potential -55 V.-this equals final p-intial p
- =-55V-(135V)
- =-190V * 1.6 * 10^-19 C
= -3.0 * 10^-17 J
2. 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.
- the change of the proton is equal to the magnitude of that of an electron so the work in the electron would be 180volts * charge of one of the electrons
=180eV
4. Potential difference required to give He nucleus 65.0 keV of KE.
-65.0 keV is 65.0 * 10^3 eV, or 6.50 * 10^4 eV, of energy.
- since Helium has a nucleus of 2+ then the nucleus would gain 2eV . In order to gain 6.50 * 10^4 eV of energy the difference if half of it
5. 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?
- V=kq/r
-q = 1.60*10^-19C..
-9.0*10^9N*m^2/C^2(1.60*10^-19C) / (2.5*10^-15m)
= 5.8*10^5V.
- The potential energy would have to be calculated using two protons which would be PE=(1.60*10^-19C)(5.8*10^5V)
= 9.2*10^-14 J.
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