Find the surface charge density at the inters | vedclass

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Question

(A) The electric field $E$ due to an infinite line charge with linear charge density $\lambda$ at a radial distance $r$ is given by $E = \frac{\lambda

Vedclass · Accepted Answer

$0.424 \, nC/m^2$

Vedclass · Answer

(A) The electric field $E$ due to an infinite line charge with linear charge density $\lambda$ at a radial distance $r$ is given by $E = \frac{\lambda}{2 \pi \varepsilon_0 r}$.Given $\lambda = 8 	imes 10^{-9} \, C/m$ and $r = x = 3 \, m$.The electric field at $x = 3 \, m$ is $E = \frac{8 	imes 10^{-9}}{2 \pi \varepsilon_0 (3)}$.Using the boundary condition for the surface charge density $\sigma$ at a conductor surface,$E = \frac{\sigma}{\varepsilon_0}$,we have $\sigma = E \varepsilon_0$.Substituting $E$,we get $\sigma = \frac{\lambda}{2 \pi r} = \frac{8 	imes 10^{-9}}{2 \pi (3)}$.$\sigma = \frac{8 	imes 10^{-9}}{6 \pi} \approx 0.424 	imes 10^{-9} \, C/m^2 = 0.424 \, nC/m^2$.

Find the surface charge density at the intersection of the plane $x = 3 \, m$ and the $x$-axis,given a uniform line charge of $8 \, nC/m$ lying along the $z$-axis in free space.

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