Question:
A disk of radius a/4 having a uniformly distributed charge -6C is placed in the x-y plane with its centre at (-a/2,0,0). A rod of length a carrying a uniformly distributed charge 8C is placed on the x-axis from x = a/4 to x = 5a/4. Two point charges -7C and 3C are placed at (-a/4,0,0) and (-3a/4,3a/4,0), respectively. Consider a cubical surface formed by six surfaces x=± a/2, y=± a/2, z=± a/2. The electric flux through this cubical surface is
A disk of radius a/4 having a uniformly distributed charge -6C is placed in the x-y plane with its centre at (-a/2,0,0). A rod of length a carrying a uniformly distributed charge 8C is placed on the x-axis from x = a/4 to x = 5a/4. Two point charges -7C and 3C are placed at (-a/4,0,0) and (-3a/4,3a/4,0), respectively. Consider a cubical surface formed by six surfaces x=± a/2, y=± a/2, z=± a/2. The electric flux through this cubical surface is
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For flux calculations, only the net enclosed charge matters — distribution and position inside do not affect the total flux.
Updated On: Mar 19, 2026
- -(2C)/(varepsilon₀)
- (2C)/(varepsilon₀)
- (10C)/(varepsilon₀)
- (12C)/(varepsilon₀)
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The Correct Option is B
Solution and Explanation
Step 1: By Gauss’s law: Φ = fracQₑₙclₒₛₑdvarepsilon₀
Step 2: Charges enclosed by the cube:
• Part of the uniformly charged rod lies inside the cube ⟹ +2C
• The disk lies completely outside the cube ⟹ 0
• One point charge inside contributes accordingly
• Net enclosed charge = 2C
Step 3: Electric flux: Φ = (2C)/(varepsilon₀)
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