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Question

(A) The electric field $E$ due to a large charged plane with surface charge density $\sigma$ is given by $E = \frac{\sigma}{2 \varepsilon_0}$.Given $\

Vedclass · Accepted Answer

$43.56 	ext{ N m}^2 	ext{ C}^{-1}$

Vedclass · Answer

(A) The electric field $E$ due to a large charged plane with surface charge density $\sigma$ is given by $E = \frac{\sigma}{2 \varepsilon_0}$.Given $\sigma = 4.9 	imes 10^{-6} 	ext{ C m}^{-2}$.The electric flux $\phi$ through an area $A$ is given by $\phi = \vec{E} \cdot \vec{A} = EA \cos 	heta$,where $	heta$ is the angle between the electric field vector and the area vector (normal to the surface).Here,the electric field is along the $z$-axis (perpendicular to the $x-y$ plane). The normal to the circular plane makes an angle $	heta = 60^{\circ}$ with the $z$-axis.Area $A = \pi r^2 = \pi 	imes (0.01 	ext{ m})^2 = \pi 	imes 10^{-4} 	ext{ m}^2$.Using $\frac{1}{4 \pi \varepsilon_0} = 9 	imes 10^9 	ext{ N m}^2 	ext{ C}^{-2}$,we have $\frac{1}{2 \varepsilon_0} = 2 \pi 	imes 9 	imes 10^9 = 18 \pi 	imes 10^9$.Thus,$\phi = \left( \frac{\sigma}{2 \varepsilon_0} ight) A \cos 60^{\circ} = (\sigma 	imes 18 \pi 	imes 10^9) 	imes (\pi 	imes 10^{-4}) 	imes \frac{1}{2}$.Substituting the values: $\phi = (4.9 	imes 10^{-6}) 	imes (18 \pi^2 	imes 10^5) 	imes 0.5 = 4.9 	imes 9 	imes \pi^2 	imes 10^{-1} = 44.1 	imes 9.8696 	imes 0.1 \approx 43.56 	ext{ N m}^2 	ext{ C}^{-1}$.

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