- #1
Ajwrighter
- 42
- 0
1. A nonconducting rod of length L = 6cm and uniform linear charge density A = +3.68pC/M . Take V = 0 at infinity. What is V at point P at distance d = 8.0cm along the rod's perpendicular bisector?
2. V = S E * ds One half of the rod = L/2 1/4piEo = 9x10^9
R = sqrt((L/2)^2 + (D^2)) = 0.08544m
3. I've attempted this solution many times but here is the more recent.
(9x10^9) [(((3.68 x 10^-12)/0.08544 ) * (.03)*(2) ] = .0232584V
My reasoning for this method. Take the Charge density A and divide it by the Range of both sides, then multiply by the length of each side which equals .03 and then multiply it by twice to obtain V from both sides (since both sides are identical.) The problem is the answer in the back gives 24.3mV = .0243V. The method I am using, is it just a close proximity, a lucky guess? Or can my method be reproduced on similar problems to obtain a close proximity? In either case, if my method is wrong what do I need to change?
2. V = S E * ds One half of the rod = L/2 1/4piEo = 9x10^9
R = sqrt((L/2)^2 + (D^2)) = 0.08544m
3. I've attempted this solution many times but here is the more recent.
(9x10^9) [(((3.68 x 10^-12)/0.08544 ) * (.03)*(2) ] = .0232584V
My reasoning for this method. Take the Charge density A and divide it by the Range of both sides, then multiply by the length of each side which equals .03 and then multiply it by twice to obtain V from both sides (since both sides are identical.) The problem is the answer in the back gives 24.3mV = .0243V. The method I am using, is it just a close proximity, a lucky guess? Or can my method be reproduced on similar problems to obtain a close proximity? In either case, if my method is wrong what do I need to change?