The figure above shows two Gaussian surfaces: a cube with side length \(d\) and a sphere with diameter \(d\). The net electric charge enclosed within each surface is the same, \(+Q\). If \( \Phi_C \) denotes the total electric flux through the cubical surface, and \( \Phi_S \) denotes the total electric flux through the spherical surface, which of the following is true?
Correct Answer: \( \boxed{\mathrm{C}} \)
Gauss’s law states that the total electric flux through any closed surface is
\( \Phi=\dfrac{Q_{\mathrm{enc}}}{\varepsilon_0} \)
where \(Q_{\mathrm{enc}}\) is the net charge enclosed by the surface.
Both the cube and the sphere enclose the same net charge, \(+Q\). Therefore, the total electric flux through each closed surface is
\( \Phi_C=\dfrac{Q}{\varepsilon_0} \)
and
\( \Phi_S=\dfrac{Q}{\varepsilon_0} \)
The shape or size of a closed Gaussian surface does not affect the total electric flux. Only the enclosed net charge determines the flux.
Hence,
\( \boxed{\Phi_C=\Phi_S} \)
Therefore, the correct answer is (C).
A uniform electric field exists in a region, and then a neutral, conducting, spherical shell with a stationary charge \(+2Q\) at its center is placed in the region, as shown above. The radius of the sphere is \(R\). The electric flux through the sphere depends on the value of
Correct Answer: \( \boxed{\mathrm{E}} \)
By Gauss’s law, the net electric flux through any closed surface is
\( \Phi_E=\oint \vec{E}\cdot d\vec{A}=\dfrac{Q_{\mathrm{enc}}}{\varepsilon_0}. \)
The conducting spherical shell encloses only the stationary charge \(+2Q\). The external uniform electric field contributes zero net flux because every field line entering the closed surface also leaves it.
Therefore,
\( \Phi_E=\dfrac{2Q}{\varepsilon_0}. \)
The flux is independent of the sphere’s radius \(R\) and the external field magnitude \(E\). It depends only on the enclosed charge, which is proportional to \(Q\).
Therefore, the correct answer is (E).
A closed surface, in the shape of a cube of side \(a\), is oriented as shown above in a region where there is a constant electric field of magnitude \(E\) parallel to the \(x\)-axis. The total electric flux through the cubical surface is
Correct Answer: \( \boxed{\mathrm{B}} \)
According to Gauss’s law,
\( \Phi_E=\oint \vec{E}\cdot d\vec{A}=\dfrac{Q_{\mathrm{enc}}}{\varepsilon_0}. \)
The cube is placed in a uniform electric field and encloses no charge.
The electric flux entering one face of the cube is exactly equal in magnitude and opposite in sign to the flux leaving the opposite face.
The remaining four faces have area vectors perpendicular to the electric field, so their individual fluxes are zero because
\( \vec{E}\cdot d\vec{A}=EA\cos90^\circ=0. \)
Therefore, the net electric flux through the entire closed surface is
\( \boxed{\Phi_E=0}. \)
Therefore, the correct answer is (B).
