Answer:
a) [tex]Q_{enc}=1.06\times 10^{-3}\ C[/tex]
b) Since no charge is enclosed by the imaginary surface at the given radial distance so the electric field measured outward form the axis is also zero.
Explanation:
Given:
radius of the cylindrical shell, [tex]r=0.07\ m[/tex]
length of the cylindrical shell, [tex]h=2.6\ m[/tex]
magnitude of electric field at 20 cm from the axis, [tex]E=36000\ N.C^{-1}[/tex]
a)
[tex]\oint E.dA=\frac{Q_{enc}}{\epsilon_0}[/tex]
[tex]\Rightarrow \vert E\vert \times 2\pi.r'.h=\frac{Q_{enc}}{\epsilon_0}[/tex]
[tex]36000\times 2\pi\times 0.2\times 2.6\times 9\times 10^{-9}=Q_{enc}[/tex]
[tex]Q_{enc}=1.06\times 10^{-3}\ C[/tex]
b)
Given radial distance, [tex]r=0.04\ m[/tex]
[tex]\oint E.dA=\frac{Q_{enc}}{\epsilon_0}[/tex]
[tex]\Rightarrow \vert E\vert \times 2\pi.r'.h=\frac{Q_{enc}}{\epsilon_0}[/tex]
[tex]\frac{Q_{enc}}{\epsilon_0} =\vert E\vert\times 2\pi.r'.h[/tex]
[tex]\frac{Q_{enc}}{\esilon_0} =0[/tex]
Since no charge is enclosed by the imaginary surface at the given radial distance so the electric field measured outward form the axis is also zero.