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Potential of a Finite Rod ( will rate for a life saver!)

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Scholar
Karma Points: 225
Respect (75%):
Date Posted: 6/30/2008 10:25:38 AM  Status: Closed
Potential of a Finite Rod ( will rate for a life saver!)
Course Textbook Chapter Problem
Calculus Based Physics University Physics with Modern Physics (12th) by Young, Freedman 24 N/A
Question Details:
A finite rod of length L has total charge q, distributed uniformly along its length. The rod lies on the x -axis and is centered at the origin. Thus one endpoint is located at (-L/2, 0), and the other is located at (L/2, 0). Define the electric potential to be zero at an infinite distance away from the rod. Throughout this problem, you may use the constant k in place of the expression \frac{1}{4\pi\epsilon_0}.


Part A
 (Part A 1 figure) 
What is V_A, the electric potential at point A (see the figure), located a distance d above the midpoint of the rod on the y axis?
Express your answer in terms of L, d, q, and k.
  V_A  =




Part B

What is V_B, the electric potential at point B, located at distance d from one end of the rod (on the x axis)?

 (Part B 1 figure) 
Give your answer in terms of q, L, d, and k.
  V_B  =



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Answers:

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Oracle
Karma Points: 15,230
Date Posted: 7/2/2008 5:13:49 PM  Status: Live
Asker's Rating: Lifesaver   
Response:
Part A
dq = (q/L) dx
dV = k dq/r
r = √(x2 + d2)
dV = (kq/L) dx/√(x2 + d2)
integrate, limits: -L/2 to L/2
VA = (kq/L) ln[√(x2 + d2) + x] = (plug limits) = (kq/L) {ln[√((L/2)2 + d2) + L/2] - ln[√((L/2)2 + d2) - L/2]}
sadgirl's Comment:
that's great answer thanks !

Member's Avatar
Oracle
Karma Points: 15,230
Date Posted: 7/2/2008 5:16:58 PM  Status: Live
Asker's Rating: Helpful   
Response:
Part B
dq = (q/L) dr
dV = k dq/r = (kq/L) dr/r
integrate, limits: d to d + L
VB = (kq/L) ln r = (plug limits) = (kq/L) ln[(d + L)/d}
sadgirl's Comment:
i like the way how u explain the answers ! thank you!



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