Consider the electric field created by a very long charged line of negative linear charge density -2.50 nC/m.
A small positive point charge of 8 mC moves from a distance of 9 cm to a distance of 17 cm.
How much work is done by the electric field?
Hint: The electric field for a long charged line is:
Express the result in the unit mJ and to three significant figures.
5 answers
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
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
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.)
F of Electric field----> <-- + F of Positive charge
(If you did negative work, the positive charge would move in the direction of the field.)
1 answer