A long thin wire with charge per unit length + lambda is embedded within a long

Question

A long thin wire with charge per unit length + lambda is embedded within a long cylinder of negatively charged insulating material with charge per unit volume - rho. The cylinder has radius R and the wire is sufficiently thin that you can ignore its radius. The figure above shows the cross section of the wire and cylinder and a cutaway showing the wire embedded inside the cylinder in 3D. The cylinder is the same length as the wire. a) What is the total charge enclosed, q_in, within a Gaussian cylinder with radius r (r r_ . g) If the proton is released from a long ways away from the cylinder and wire with no other forces acting on it, describe in qualitative terms how it will behave due to the forces that the cylinder and wire exert on it. In what regions is it attracted to the cylinder? Where is the force zero? Where is it repelled from the cylinder? You can assume that the proton is free to move within the insulator because it is so small.

Explanation / Answer

given,. charge per unit length in wire = lambda

charge per unit volume in cylinder = -rho

radius of cylinder = R

a. consider a gaussean surface at r < R

length of gaussean surface cylinder = L

so enclosed charge qen = (lambda*L - rho*pi*r^2*L) = L(lambda - pi*r^2*rho)

b. as the enclosing charge is -ve and the charge on wire is +ve, the qenc has a sign that depends on r

so when qenc = 0

lambda = pi*r^2*rho

r = sqroot(lambda/pi*rho)

c. From gauss' law

E*2*pi*r*L = qenc/epsilon [ where epsilon is permittivity of free space]

so for r < R

E = (lambda - pi*r^2*rho)/2*pi*epsilon

d. for R > sqroot(lambda/pi*rho)

there is no value of r for which E will be 0, except for r -> infinity

e. for r > R

qenc = (lambda*L - rho*pi*R^2*L)

so E*2*pi*r*L = (lambda*L - rho*pi*R^2*L)/epsilon

E = (lambda - rho*pi*R^2)/epsilon**2*pi*r

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