An infinite,uniformly charged sheet with surf | vedclass

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(D) According to Gauss's law,the total electric flux $\Phi$ through a closed surface is given by $\Phi = \frac{Q_{	ext{enclosed}}}{\epsilon_0}$.The i

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$\frac{{\pi {{\left( {{R^2} - {x^2}} \right)}^{}}\sigma }}{{{\varepsilon _0}}}$

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(D) According to Gauss's law,the total electric flux $\Phi$ through a closed surface is given by $\Phi = \frac{Q_{	ext{enclosed}}}{\epsilon_0}$.The infinite charged sheet intersects the spherical Gaussian surface,creating a circular cross-section of charge inside the sphere.From the geometry shown in the figure,the radius $a$ of this circular cross-section is related to the sphere radius $R$ and distance $x$ by the Pythagorean theorem: $a^2 + x^2 = R^2$,which gives $a^2 = R^2 - x^2$.The area of this circular cross-section is $A = \pi a^2 = \pi(R^2 - x^2)$.The charge enclosed by the Gaussian surface is $Q_{	ext{enclosed}} = \sigma A = \sigma \pi(R^2 - x^2)$.Substituting this into Gauss's law,the electric flux is $\Phi = \frac{\sigma \pi(R^2 - x^2)}{\epsilon_0}$.

An infinite,uniformly charged sheet with surface charge density $\sigma$ cuts through a spherical Gaussian surface of radius $R$ at a distance $x$ from its center,as shown in the figure. The electric flux $\Phi$ through the Gaussian surface is

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