A long copper cylindrical shell of inner radius 3 cm and outer radius 9 cm surrounds concentrically a charged long aluminum rod of radius 1 cm with a charge density of 7 pC/m. All charges on the aluminum rod reside at its surface. The inner surface of the copper shell has exactly opposite charge to that of the aluminum rod while the outer surface of the copper shell has the same charge as the aluminum rod. Find the magnitude and direction of the electric field (in N/C) at points that are at the following distances from the center of the aluminum rod. (If the magnitude is zero, select "none" for the direction.)

Find the magnitude and direction of the electric field at points that are at the following distances from the center of the aluminum rod: (a) 0.5 cm, (b) 1.5 cm, (c) 2.5 cm, (d) 3.5 cm, and (e) 7 cm.

A) E=0

B) E = 8.4 N/C

C) E = 0

D) E = 3.6 N/C

E) E = 1.8 N/C

Explanation:

Charge density λ = 7 x 10^-12 C/m

Electric force constant = 1/(4πεo) = 9.0 x 10^9 N • m^2 / C^2.

A) at a distance of 0.5cm from the centre of the aluminum rod, we have a conducting rod and so the charge will spread out on the surface of the conductor which means that there will be no electric field inside the rod.

So qenc = 0. So E=0

B) at a distance of 1.5cm from the centre of the aluminum rod, the point will lie outside the rod.

So, qenc = λL

So; Φ = ∮ EdA = E(2πrL)

From gausses law, Φ = qenc/εo

So, E(2πrL) = λL/εo

E = λL/(εo(2πrL)) = 2λL/(εo(4πrL))

Therefore, E = {2 x 9.0 x (10^9) x 7 x 10^(-12)}/ 0.015 = 8.4 N/C

C) At a distance of 2.5cm from the centre of the aluminum rod, the point lies inside the material of the copper cylinder.

We have a conducting cylinder and as such, the total charge inside the gaussian cylinder surface is zero since the positive charge on the aluminum rod induces an equal in magnitude negative charge on the inner surface of the copper cylinder. Thus, qenc = 0 and as such E=0.

D) At a distance 3.5cm from the center of the aluminum rod;

Using the equation for E;

We get; E = {2 x 9.0 x (10^9) x 7 x 10^(-12)}/ 0.035 = 3.6 N/C

E) At a distance 7 cm from the center of the aluminum rod;

Using the equation for E;

We get; E = {2 x 9.0 x (10^9) x 7 x 10^(-12)}/ 0.007 = 1.8 N/C

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