The electric flux passing through the cube for the given arrangement of charges placed at the corners of the cube (as shown in the figure) is
The electric flux passing through the cube for the given arrangement of charges placed at the corners of the cube (as shown in the figure) is
- A
$\phi = \frac{1}{{2{ \in _0}}}$
- B
$\phi = \frac{{ - 1}}{{2{ \in _0}}}$
- C
$\phi = \frac{{ - 1}}{{{ \in _0}}}$
- D
$\phi = \frac{1}{{{ \in _0}}}$
Similar Questions
An arbitrary surface encloses a dipole. What is the electric flux through this surface ?
An arbitrary surface encloses a dipole. What is the electric flux through this surface ?
The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about $150\, N/C$, directed inward towards the center of the Earth . This gives the total net surface charge carried by the Earth to be......$kC$ [Given ${\varepsilon _0} = 8.85 \times {10^{ - 12}}\,{C^2}/N - {m^2},{R_E} = 6.37 \times {10^6}\,m$]
The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about $150\, N/C$, directed inward towards the center of the Earth . This gives the total net surface charge carried by the Earth to be......$kC$ [Given ${\varepsilon _0} = 8.85 \times {10^{ - 12}}\,{C^2}/N - {m^2},{R_E} = 6.37 \times {10^6}\,m$]
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The black shapes in the figure below are closed surfaces. The electric field lines are in red. For which case, the net flux through the surfaces is non-zero?
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The flat base of a hemisphere of radius $a$ with no charge inside it lies in a horizontal plane. A uniform electric field $\vec E$ is applied at an angle $\frac {\pi }{4}$ with the vertical direction. The electric flux through the curved surface of the hemisphere is
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