Q11.A cube is placed inside an electric field, →𝐸= 150𝑦2^j The side of the cube is 0 . 5 m and is placed in the field as shown in the given figure. The charge inside the cube is: (1) 8 . 3 × 10-11C (2) 3 . 8 × 10-11C (3) 3 . 8 × 10-12C (4) 8 . 3 × 10-12C
What This Question Tests
This problem applies Gauss's Law to find the charge enclosed within a cube in a non-uniform electric field. It requires calculating the electric flux through each face of the cube and then summing them up to find the total flux, which is proportional to the enclosed charge.
Concepts Tested
Formulas Used
Φ = ∫ E ⋅ dA
Φ = Q_enclosed / ϵ_0
📚 NCERT Sections This Tests
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?
1.15 — 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?
1.14 — 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?
📋 Question Details
- Chapter
- Electrostatics
- Topic
- Gauss's Law
- Year
- 2021
- Shift
- 01 Sep Shift 2
- Q Number
- Q11
- Type
- MCQ
- NCERT Ref
- Class 12 Physics Ch 1: Electric Charges and Fields
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