dillonmhudson
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Any answers would be great, thanks!
1. A charge Q is uniformly spread along the x-axis from x=0 to x=l. Find the electric potential function at points on the x-axis for x>l.
2. An electron moves from one point to another where the second point has a larger value of the electric potential by 5 volts. If the initial velocity was zero, how fast will the electron be going at the second point? (The mass of the electron is 9.11E-31 kg)
3. Suppose an electron were moving in a straight line towards a fixed nucleus which had a positive charge 8 times the size of the charge on an electron. If the electron started at infinity with essentially zero velocity, how fast would it be moving when it was 50E-11 m from the nucleus?
Coulomb's Law = \alpha*[q1*q2]/r^2
where \alpha=1/[4*\pi*\epsilon_{}0
U=1/[4*pi*epsilon-sub0]*[q*q sub0]/r
V=U/[q sub0]
1. dV=1/[4*pi*epsilon]*[(Q/l)*dx]/[x-l]
dV=1/[4*pi*epsilon]*Q/l
I don't know what to do with this one, the answer has a natural log (ln) in it.
2. KE(initial) + U(initial) = KE(final) + U(final)
0+1/[4*pi*epsilon]*q/r = 1/2 * m * v(final)^2 + 1/[4*pi*epsilon]*q/r
Not sure what to do here either, seem to have three unknowns
Answer = 1.33E6 m/s
3. KE(initial) + U(initial) = KE(final) + U(final)
0=1/2*9.109E-31(mass of an electron)*v(final)^2+9E9*[8*(1.6E-19)^2]/[50E-11]
I don't understand this one eitherThanks for all the help!
Homework Statement
1. A charge Q is uniformly spread along the x-axis from x=0 to x=l. Find the electric potential function at points on the x-axis for x>l.
2. An electron moves from one point to another where the second point has a larger value of the electric potential by 5 volts. If the initial velocity was zero, how fast will the electron be going at the second point? (The mass of the electron is 9.11E-31 kg)
3. Suppose an electron were moving in a straight line towards a fixed nucleus which had a positive charge 8 times the size of the charge on an electron. If the electron started at infinity with essentially zero velocity, how fast would it be moving when it was 50E-11 m from the nucleus?
Homework Equations
Coulomb's Law = \alpha*[q1*q2]/r^2
where \alpha=1/[4*\pi*\epsilon_{}0
U=1/[4*pi*epsilon-sub0]*[q*q sub0]/r
V=U/[q sub0]
The Attempt at a Solution
1. dV=1/[4*pi*epsilon]*[(Q/l)*dx]/[x-l]
dV=1/[4*pi*epsilon]*Q/l
I don't know what to do with this one, the answer has a natural log (ln) in it.
2. KE(initial) + U(initial) = KE(final) + U(final)
0+1/[4*pi*epsilon]*q/r = 1/2 * m * v(final)^2 + 1/[4*pi*epsilon]*q/r
Not sure what to do here either, seem to have three unknowns
Answer = 1.33E6 m/s
3. KE(initial) + U(initial) = KE(final) + U(final)
0=1/2*9.109E-31(mass of an electron)*v(final)^2+9E9*[8*(1.6E-19)^2]/[50E-11]
I don't understand this one eitherThanks for all the help!
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