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If the net charge enclosed by a closed Gaussian surface is zero, does this mean that the electric field at all points on the surface is also zero?


Yes; because $\oint \overrightarrow{E}\cdot d\overrightarrow{A} = \dfrac{Q_{enc}}{\epsilon_0}$, if $Q_{enc}$ is zero, then $\overrightarrow{E}$ must be zero.


Yes; electric field strength depends on how much net charge is present in a region.


Yes; electric field can be cancelled out by charges outside the surface.


No; there may be electric field that causes positive and negative flux at different points.


No; there is no relationship between charge and electric field.

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