Answer:
(a) E=233.56 N/C
(b) The surface charge density of inner surface σ= -5.69×10⁻⁹C/m²
(c)The surface charge density of outer surface σ= 3.25×10⁻⁹C/m²
Explanation:
For Part (a)
The magnitude of the electric field at distance of 15.4 cm from the axis of the shell is given by:
E=λ/2πε₀r
Substitute the given values
[tex]E=\frac{2.00*10^{-9} }{2\pi 8.854*10^{-12}*0.154 }\\ E=233.56N/C[/tex]
Since the nonconducting rod positively charged,it induces a negative charge -q on the inner surface of conducting shell and a positive charge +q on the outer surface of conducting shell,so the net charge of conducting shell is zero
Part (b)
The surface charge density of inner surface is given by:
σ=-q/A
[tex]=\frac{-q}{2\pi r_{inner} l}\\ =\frac{\frac{-q}{l} }{2\pi r_{inner}}\\[/tex]
= -λ/2πr
[tex]=\frac{-2.0*10^{-9} }{2\pi (0.056)}\\ =-5.69*10^{-9}C/m^{2}[/tex]
Part(c)
Similarly the surface charge density on the outer surface of the cylindrical shell given by:
σ=λ/2πr
[tex]=\frac{2.0*10^{-9} }{2\pi 0.098}\\ =3.25*10^{-9}C/m^{2}[/tex]