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(A) The electric flux $\phi$ through the curved surface of a cylinder of radius $r$ and length $L$ is given by $\phi = E 	imes A$,where $A$ is the cu

Vedclass · Accepted Answer

$1.1 	imes 10^2$

Vedclass · Answer

(A) The electric flux $\phi$ through the curved surface of a cylinder of radius $r$ and length $L$ is given by $\phi = E 	imes A$,where $A$ is the curved surface area.The curved surface area $A = 2 \pi r L$.Given: $E = 250 \,NC^{-1}$,$r = 7 \,cm = 0.07 \,m$,and $L = 1 \,m$.Substituting the values:$\phi = 250 	imes (2 	imes \pi 	imes 0.07 	imes 1)$$\phi = 250 	imes (0.14 \pi)$$\phi = 35 \pi$Using $\pi \approx 3.14159$:$\phi \approx 35 	imes 3.14159 \approx 109.95 \,Nm^2C^{-1}$Rounding to two significant figures,we get $\phi \approx 1.1 	imes 10^2 \,Nm^2C^{-1}$.

An infinite line charge is at the axis of a cylinder of length $1 \,m$ and radius $7 \,cm$. If the electric field at any point on the curved surface of the cylinder is $250 \,NC^{-1}$,then the net electric flux through the cylinder is ............ $Nm^2C^{-1}$.

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