Question:
A charge 'Q' C is placed at the centre of a cube. The electric flux through two opposite faces of the cube is
(\(ε_0\) = permittivity of free space)
A charge 'Q' C is placed at the centre of a cube. The electric flux through two opposite faces of the cube is
(\(ε_0\) = permittivity of free space)
(\(ε_0\) = permittivity of free space)
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Gauss law gives the total flux Q over epsilon naught, and symmetry splits it equally over the six faces.
Updated On: Oct 1, 2026
- \(\frac{Q}{6ε_0}\)
- \(\frac{Q}{3ε_0}\)
- \(\frac{Q}{ε_0}\)
- \(\frac{Q}{2ε_0}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
By Gauss law, the total flux through a closed surface is \(\dfrac{Q}{\varepsilon_0}\). A charge at the centre of a cube is symmetric with respect to all faces.
Step 2: Flux per face.
The cube has 6 identical faces, so each gets \(\dfrac{Q}{6\varepsilon_0}\).
Step 3: Two opposite faces.
\[ 2\times\frac{Q}{6\varepsilon_0} = \frac{Q}{3\varepsilon_0} \]
Step 4: Check the options.
\(\dfrac{Q}{6\varepsilon_0}\) is one face, \(\dfrac{Q}{\varepsilon_0}\) is the whole cube and \(\dfrac{Q}{2\varepsilon_0}\) is three faces.
Final Answer:
The flux through two opposite faces is \(\dfrac{Q}{3\varepsilon_0}\), option (B). \[ \boxed{\frac{Q}{3\varepsilon_0}} \]
By Gauss law, the total flux through a closed surface is \(\dfrac{Q}{\varepsilon_0}\). A charge at the centre of a cube is symmetric with respect to all faces.
Step 2: Flux per face.
The cube has 6 identical faces, so each gets \(\dfrac{Q}{6\varepsilon_0}\).
Step 3: Two opposite faces.
\[ 2\times\frac{Q}{6\varepsilon_0} = \frac{Q}{3\varepsilon_0} \]
Step 4: Check the options.
\(\dfrac{Q}{6\varepsilon_0}\) is one face, \(\dfrac{Q}{\varepsilon_0}\) is the whole cube and \(\dfrac{Q}{2\varepsilon_0}\) is three faces.
Final Answer:
The flux through two opposite faces is \(\dfrac{Q}{3\varepsilon_0}\), option (B). \[ \boxed{\frac{Q}{3\varepsilon_0}} \]
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