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PhysicsMediumNumerical2021 · 24 Feb Shift 2

Q26.A point charge of +12 μC is at a distance 6 cm vertically above the centre of a square of side 12 cm as shown in figure. The magnitude of the electric flux through the square will be _______ ×103 N m2 C−1. JEE Main 2021 (24 Feb Shift 2) JEE Main Previous Year Paper

What This Question Tests

This question tests the application of Gauss's Law and symmetry to find the electric flux through a specific planar surface due to a point charge.

Concepts Tested

Gauss's LawElectric flux through a surfaceSymmetry considerations

Formulas Used

Φ = Q_enclosed / ε₀

📚 NCERT Sections This Tests

1.17A Point Charge +10 Mc Is A Distance 5 Cm Directly Above The Centre

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1.17 A point charge +10 mC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.31. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.) FIGURE 1.31

1.14Consider A Uniform Electric Field E = 3 × 103 Î N/C. (A) What Is The

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1.14 Consider a uniform electric field E = 3 × 103 î N/C. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?

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1.18 A point charge of 2.0 mC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?