Question

Consider a long, thin, plastic cylindrical shell centered at the origin. It has a radius 2R and a linear charge density -3 λ .

b. Use Gauss's Law to find the electric field (mag and direction) at a distance x=3R from the origin.

Now a long line of charge (parallel to the axis of the cylinder) is added at a distance x=4R with linear charge density + λ .

c. use superposition to find the force (mag. and direction) on an electron placed at x=3R

Answer 1

part b:

let electric field be E.

let length of the cylinder be L.

using Gauss' law:

epsilon*E*2*pi*x*L=total charge enclosed=linear charge density*L

==>epsilon*E*2*pi*(3*R)=-3*lambda

==>E=-lambda/(2*pi*epsilon*R)

negative sign indicates the force is radially inward and hence towards -ve x axis.

part c:

field due to line charge=linear charge density/(2*pi*epsilon*distance)

so field due to the linear charge density lambda at x=3*R is given by

E=lambda/(2*pi*epsilon*(4*R-3*R))

=lambda/(2*pi*epsilon*R)

field is directed towards negative x axis

so total field is along negative x axis with a magnitude of

(lambda/(2*pi*epsilon*R))+(lambda/(2*pi*epsilon*R))

=lambda/(pi*epsilon*R)

force on electron=charge * electric field

magnitude of force=e*electric field

where e=1.6*10^(-19) C

hence magnitude of force=e*lambda/(pi*epsilon*R)

as charge on electron is negative, direction of force is in opposite direction to that of the field.

hence force is directed along +ve x axis