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  • (a) State Gauss’ law. Using this law, obtain the expression for the electric field due to an infinitely long straight conductor of linear charge density .<b

(a) State Gauss’ law. Using this law, obtain the expression for the electric field due
to an infinitely long straight conductor of linear charge density \lambda.
(b) A wire AB of length L has linear charge density \lambda = kx, where x is measured from the end A of the wire. This wire is enclosed by a Gaussian hollow surface. Find the expression for the electric flux through this surface.

 

 

 
 
 
 
 

Answers (1)

a) Gauss' Law
The total electric flux over a closed surface is equal to \frac{1}{\varepsilon _{0}} times the net charge enclosed within the surface
 \phi _{E}= \oint E\cdot ds= \frac{1}{\varepsilon _{0}}q
Consider an infinity long straight conductor of linear charge density \lambda

Consider a cylindrical Gaussian surface of radius r and length l.
charge enclosed by the surface
q= \lambda l
By Gauss law
\oint \vec E\cdot ds= \frac{q}{\varepsilon _{0}}= \frac{\lambda l}{\varepsilon _{0}}
\vec E\oint ds= \frac{\lambda l}{\varepsilon _{0}}
\vec E\times 2\pi rL= \frac{\lambda L}{\varepsilon _{0}}
E= \frac{\lambda }{2\pi r\varepsilon _{0}}
b)
given linear charge density \lambda = kx
consider a small length dx of the wire, the charge
dq= \lambda dx
The total charge of the wire AB is
q= \int_{0}^{L}\lambda xdx= \frac{\lambda L^{2}}{2}
By Gauss law
flux \phi = \oint E\cdot ds= \frac{q}{\varepsilon _{0}}= \frac{\lambda L^{2}}{2\varepsilon _{0}}

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Safeer PP

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