- #1
Trenthan
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Ey all
Im a little confussed with electric fields and potential. My textbook says one thing and my tutor has said the opposite*, so I am not sure what to belive.
If we have a "uniformly" charged necleus(we can model as a uniformly charged sphere) thus the charge(protons) are spread throughout the whole volume of the sphere not just the surface** (only at surface when all points in the sphere are in electrostatic equilibrium which isn't stated*)
Therefore the electric field would be
R- radius of necleus, r- radius,
Eoutside necleus = (1/(4*pi*e0))*(Q/r2), When r>R
Einside necleus = (1/(4*pi*e0))*(Q*r/R3),When r<R
My totor said that E=0 inside the nucleus, I am hoping i only copied down what she said wrong :S, if someone can confirm please. If the formula are correct therefore the elctric field increases as we get closer to the center of the nucleus
Electrostatic potential,(doesnt state that neucleus is in electrostatic equilibrium so I am unsure why they don't simply ask for potential* anyway, if someone can explain i would appreciate it, from what I've read they are the same)
Voutside necleus = (1/(4*pi*e0))*(Q/r) When r>R
Vinside necleus = (1/(4*pi*e0))*(Q/R) When r<R
Assuming the formula are correct therefore the potential inside the nucleus is constant? and decreases by 1/r outside the nucleus,
Cheers Trent
Im a little confussed with electric fields and potential. My textbook says one thing and my tutor has said the opposite*, so I am not sure what to belive.
If we have a "uniformly" charged necleus(we can model as a uniformly charged sphere) thus the charge(protons) are spread throughout the whole volume of the sphere not just the surface** (only at surface when all points in the sphere are in electrostatic equilibrium which isn't stated*)
Therefore the electric field would be
R- radius of necleus, r- radius,
Eoutside necleus = (1/(4*pi*e0))*(Q/r2), When r>R
Einside necleus = (1/(4*pi*e0))*(Q*r/R3),When r<R
My totor said that E=0 inside the nucleus, I am hoping i only copied down what she said wrong :S, if someone can confirm please. If the formula are correct therefore the elctric field increases as we get closer to the center of the nucleus
Electrostatic potential,(doesnt state that neucleus is in electrostatic equilibrium so I am unsure why they don't simply ask for potential* anyway, if someone can explain i would appreciate it, from what I've read they are the same)
Voutside necleus = (1/(4*pi*e0))*(Q/r) When r>R
Vinside necleus = (1/(4*pi*e0))*(Q/R) When r<R
Assuming the formula are correct therefore the potential inside the nucleus is constant? and decreases by 1/r outside the nucleus,
Cheers Trent
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