Posted by Susie on Friday, February 12, 2010 at 12:55pm.
The Equation is E line = 1/(4piE0)(2lamda/r)
Since you have not provided the "hint", I used Gauss' Law to come up with
2*pi*R*L*E = L*q/eo
where eo is the coulomb's law constant "epsilon-zero" = 8.89^10^-12 N*m^2/C^2
and q is the charge per unit length
E(R) = q/(2 pi eo R)
Multiply that by Q = 8 mC and integrate with dR from R = 0.09 to 0.17 m
There should be a ln(17/8) term in the answer.
The work done will be positive since Q is moving in the opposite direction from the attractive force
The hint is
E line = 1/(4piE0)(2lamda/r)
Are the answers the same even if I use Gauss' Law to come up with 2*pi*R*L*E = L*q/eo
Isn't the work negative, because this is a positive charge moving in the opposite direction of the electric field?
Positive work must be done in order to make a positive charge go opposite the direction of the electric field.
F of Electric field----> <-- + F of Positive charge
(If you did negative work, the positive charge would move in the direction of the field.)
Related Questions
physics - Consider the electric field created by a very long charged line of ...
Physics - Consider the electric field created by a very long charged line of ...
Physics - Two particles are fixed along a straight line, separated by a distance...
Physics - An infinitely long line charge of uniform linear charge density &...
Physics - An infinitely long line charge of uniform linear charge density &#...
PHYSICS - A charged disk of radius R that carries a surface charge density &...
Physics - A charged disk of radius R that carries a surface charge density &...
Physics; Elect. - 1. A charge (uniform linear density=9.0 nC/m) is distributed ...
Physics - Given a square of sidelength a = 4 cm. We place a charged particle at ...
physics E&M - I posted a question on here earlier but I wasn't able to ...
For Further Reading
