Q27.The total charge enclosed in an incremental volume of 2 × 10−9 m3 located at the origin is nC, if electric flux density of its field is found as D = e−x sin yˆi −e−x cosyˆj + 2zˆk C m−2
What This Question Tests
This question requires calculating the charge enclosed in a small volume using the differential form of Gauss's Law, involving the divergence of the electric flux density.
Concepts Tested
Formulas Used
∇ . D = ρ
Q_enclosed = ρ * ΔV
📚 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?
1.19 — A Point Charge Causes An Electric Flux Of –1.0 × 103 Nm2/C To Pass
Physics Class 11 · Chapter 1
1.19 A point charge causes an electric flux of –1.0 × 103 Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?
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
- Gauss's Law in differential form
- Year
- 2021
- Shift
- 22 Jul Shift 1
- Q Number
- Q27
- Type
- Numerical
- NCERT Ref
- Class 12 Physics Ch 1: Electric Charges and Fields
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