In finding the electric field using Gauss's L | vedclass

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(D) Gauss's Law in integral form is given by $\oint \overrightarrow{E} \cdot d\overrightarrow{A} = \frac{q_{enc}}{\varepsilon_{0}}$.To simplify this t

Vedclass · Accepted Answer

Only when the Gaussian surface is an equipotential surface and $|\overrightarrow{E}|$ is constant on the surface.

Vedclass · Answer

(D) Gauss's Law in integral form is given by $\oint \overrightarrow{E} \cdot d\overrightarrow{A} = \frac{q_{enc}}{\varepsilon_{0}}$.To simplify this to the form $|\overrightarrow{E}| |A| = \frac{q_{enc}}{\varepsilon_{0}}$,we must ensure that the electric field magnitude $|\overrightarrow{E}|$ is constant over the surface and that the electric field vector $\overrightarrow{E}$ is parallel to the area vector $d\overrightarrow{A}$ at every point on the surface.If $|\overrightarrow{E}|$ is constant and the surface is chosen such that $\overrightarrow{E}$ is always normal to the surface,then the surface is an equipotential surface.Therefore,both conditions are required to pull $|\overrightarrow{E}|$ out of the integral.

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