## Electric Potential Questions

Any answers would be great, thanks!

1. The problem statement, all variables and given/known data
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?

2. Relevant 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]

3. 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 either

Thanks for all the help!
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Recognitions:
Homework Help
 Quote by dillonmhudson Any answers would be great, thanks! 1. The problem statement, all variables and given/known data 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? 2. Relevant 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] 3. 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.
Let s be the position on the x axis at which the potential is measured. Write out the expression for potential due to the charge element Qdx/l at a position x where $0 \le x \le l$

 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
How much kinetic energy does the electron acquire in moving to the second point? (Hint: for a negative charge, the second position is at a lower potential energy per unit negative charge by 5 joules/coulomb).

 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 either
What is the potential energy of the electron at infinity? What is it at the given distance? Where does that potential energy go?

AM
 I'm still not getting question 4...

## Electric Potential Questions

There is no question 4.
 sorry it's 4 in my book, question 1
 The electric potential at point P, along the line, is kq/r d is the distance from the end of the rod to the point. So... $$dV = \frac{k}{x}dq$$ $$dq = \lambda dx = Q/L dx$$ $$dV = \frac{kQ}{Lx}dx$$ Then integrate from d to L + d.
 Ok thanks! Got it! I had: int[k*[(Q/L)dx]/(a-x)] from 0,L where L is the distance of where the charge is spread Thanks a lot!

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