A charge $Q$ is placed at a distance $a/2$ above the centre of the square surface of edge $a$ as shown in the figure. The electric flux through the square surface is
A charge $Q$ is placed at a distance $a/2$ above the centre of the square surface of edge $a$ as shown in the figure. The electric flux through the square surface is
- [JEE MAIN 2018]
- A
$\frac{Q}{{3{\varepsilon _0}}}$
- B
$\frac{Q}{{6{\varepsilon _0}}}$
- C
$\frac{Q}{{2{\varepsilon _0}}}$
- D
$\frac{Q}{{{\varepsilon _0}}}$
Similar Questions
Draw electric field by negative charge.
Draw electric field by negative charge.
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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 $\left.\mathrm{Nm}^{2} / \mathrm{C}\right)$
- [JEE MAIN 2020]
What is the direction of electric field intensity ?
What is the direction of electric field intensity ?
A cube is placed inside an electric field, $\overrightarrow{{E}}=150\, {y}^{2}\, \hat{{j}}$. The side of the cube is $0.5 \,{m}$ and is placed in the field as shown in the given figure. The charge inside the cube is $.....\times 10^{-11} {C}$
A cube is placed inside an electric field, $\overrightarrow{{E}}=150\, {y}^{2}\, \hat{{j}}$. The side of the cube is $0.5 \,{m}$ and is placed in the field as shown in the given figure. The charge inside the cube is $.....\times 10^{-11} {C}$
- [JEE MAIN 2021]
A point charge $q$ is placed on the centre of a hemispherical surface as shown in figure.The net flux of electric fietd tnroug the hemi-spherical surface is closest to
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