Q26.Expression for an electric field is given by →𝐸= 4000 𝑥2 ^i V m-1. The electric flux through the cube of side 20 cm when placed in electric field (as shown in the figure) is ______ V cm. JEE Main 2023 (31 Jan Shift 1) JEE Main Previous Year Paper
What This Question Tests
This question requires calculating electric flux through a cube where the electric field is non-uniform and parallel to one axis, using the definition of flux.
Concepts Tested
Formulas Used
Φ = ∫ E ⋅ dA
📚 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?
1.15 — What Is The Net Flux Of The Uniform Electric Field Of Exercise 1.14
Physics Class 11 · Chapter 1
1.15 What is the net flux of the uniform electric field of Exercise 1.14 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?
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?
📋 Question Details
- Chapter
- Electrostatics
- Topic
- Electric flux using Gauss's Law
- Year
- 2023
- Shift
- 31 Jan Shift 1
- Q Number
- Q26
- Type
- Numerical
- NCERT Ref
- Class 12 Physics Ch 1: Electrostatics
More from this Chapter
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