If $\oint_s \vec{E} \cdot \overrightarrow{d S}=0$ over a surface, then:
If $\oint_s \vec{E} \cdot \overrightarrow{d S}=0$ over a surface, then:
- [NEET 2023]
- A
the electric field inside the surface is necessarily uniform.
- B
the number of flux lines entering the surface must be equal to the number of flux lines leaving it.
- C
the magnitude of electric field on the surface is constant.
- D
all the charges must necessarily be inside the surface.
Similar Questions
Figure shows electric field lines due to a charge configuration, from this we conclude that
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- [JEE MAIN 2018]
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A hollow cylinder has a charge $q$ coulomb within it. If $\phi$ is the electric flux in units of $volt-meter$ associated with the curved surface $B,$ the flux linked with the plane surface $A$ in units of $V-m$ will be
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- [AIIMS 2008]
An electric field $\overrightarrow{\mathrm{E}}=4 \mathrm{x} \hat{\mathrm{i}}-\left(\mathrm{y}^{2}+1\right) \hat{\mathrm{j}}\; \mathrm{N} / \mathrm{C}$ passes through the box shown in figure. The flux of the electric field through surfaces $A B C D$ and $BCGF$ are marked as $\phi_{I}$ and $\phi_{\mathrm{II}}$ respectively. The difference between $\left(\phi_{\mathrm{I}}-\phi_{\mathrm{II}}\right)$ is (in $\left.\mathrm{Nm}^{2} / \mathrm{C}\right)$
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- [JEE MAIN 2020]