RankLab
Back to Questions
PhysicsMediumNumerical2023 · 01 Feb Shift 2

Q15.A cubical volume is bounded by the surfaces x = 0, x = a, y = 0, y = a, z = 0, z = a . The electric field in → the region is given by E = E0xˆi. Where E0 = 4 × 104 NC−1 m−1 . If a = 2 cm , the charge contained in the cubical volume is Q × 10–14 C . The value of Q is ______. (Take ϵ0 = 9 × 10−12 C2 N−1m−2 )

What This Question Tests

This problem requires applying Gauss's Law in integral form to find the total charge enclosed within a cubical volume given a spatially varying electric field.

Concepts Tested

Electric fluxCharge enclosedVolume integral

Formulas Used

[vec E = E_0x\hat{i}]

Φ = ∫ E ⋅ dA

Φ = Q_enclosed / ϵ₀

📚 NCERT Sections This Tests

1.18A Point Charge Of 2.0 Mc Is At The Centre Of A Cubic Gaussian

Physics Class 11 · Chapter 1

82% match

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?

1.15What Is The Net Flux Of The Uniform Electric Field Of Exercise 1.14

Physics Class 11 · Chapter 1

82% match

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?

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

Physics Class 11 · Chapter 1

81% match

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?