Q14.A charge Q is placed at a distance a/2 above the centre of a square surface of side length a. The electric flux through the square surface due to the charge would be? (1) Q (2) Q 6ϵ0 2ϵ0 (3) Q (4) Q 3ϵ0 ϵ0 JEE Main 2018 (15 Apr) JEE Main Previous Year Paper
What This Question Tests
This question requires using Gauss's Law and symmetry arguments to calculate the electric flux through a square surface due to a point charge placed centrally above it, by considering it as one face of a cube.
Concepts Tested
Formulas Used
Φ = Q_enclosed / ε₀
📚 NCERT Sections This Tests
1.14 — Consider A Uniform Electric Field E = 3 × 103 Î N/C. (A) What Is The
Physics Class 11 · Chapter 1
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?
1.18 — A Point Charge Of 2.0 Mc Is At The Centre Of A Cubic Gaussian
Physics Class 11 · Chapter 1
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?
1.17 — A Point Charge +10 Mc Is A Distance 5 Cm Directly Above The Centre
Physics Class 11 · Chapter 1
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
📋 Question Details
- Chapter
- Electrostatics
- Topic
- Electric flux (Gauss's Law)
- Year
- 2018
- Shift
- 15 Apr
- Q Number
- Q14
- Type
- MCQ
- NCERT Ref
- Class 12 Physics Ch 1: Electric Charges and Fields
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