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A long straight wire has fixed negative charge with a linear charge density of magnitude 3.8 nC/m. The wire is to be enclosed by a coaxial, thin-walled, nonconducting cylindrical shell of radius 1.6 cm. The shell is to have positive charge on its outside surface with a surface charge density σ that makes the net external electric field is zero. Calculate σ.

Answer :

opudodennis

Answer:

[tex]\sigma=3.5810\times 10^-^8C/m^2[/tex]

Explanation:

1. We  need two market [tex]\sigma[/tex] for the cylindrical shell and second for the rod.

By Gauss Law for shell, [tex]\epsilon \int E.dA=q_e_n_c[/tex] and [tex]q_e_n_c=\sigma A[/tex] then eq. becomes  [tex]\epsilon \int E.dA=A[/tex], we integrate to find

[tex]\epsilon E=\sigma[/tex]....[tex]eqtn 1[/tex]

Gauss Law for rod, [tex]\epsilon \int E.dA=\lambda L[/tex] Where [tex]\lambda[/tex] is charge per length and [tex]L[/tex] is the rod's length ([tex]q_e_n_c=\lambda L[/tex]). We integrate to find

[tex]\epsilon E=-\frac{\lamda L}{2\pi RL}.....eqtn2[/tex]

-solve the two equations to find the value of [tex]\sigma[/tex]

[tex]-\frac{\lambda}{2\pi R}=\sigma\\\frac{3.8\times 10^-^9C/m}{2\times \pi \times 1.6\times 10^-^2}\\\sigma=3.5810\times 10^-^8C/m^2[/tex]

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