Exams › JEE Main › Physics

A cube of side 0.5 m is placed in a non-uniform electric field E = 150 y² j (N/C), oriented so that the relevant faces are perpendicular to the y-axis as shown. Find the net charge enclosed within the cube.

1. 3.8 * 10⁻¹¹ C
2. 8.3 * 10⁻¹¹ C
3. 3.8 * 10⁻¹² C
4. 8.3 * 10⁻¹² C

Correct answer: 8.3 * 10⁻¹² C

Solution

The field points along +y and varies as y². Flux through the bottom (y=0) face is zero, and through the top (y=a) face is E(a)*a². With a = 0.5 m, E(a) = 150*(0.5)² = 37.5 N/C, area = 0.25 m², so flux = 9.375 N m²/C. Then q = epsilon0 * flux about 8.3 * 10⁻¹² C.

Related JEE Main Physics questions

• What is the SI unit of the permittivity of free space, ε₀?
• What is the SI unit used for electric flux?
• A constant electric field E is directed along the positive x-axis. If a charge of 0.5 C is displaced by 2 m along a line inclined at 60° to the x-axis, and the work done in this process is 10 J, what is the magnitude of the electric field?
• An electric dipole is located at the origin O along the x-axis. A point P lies 20 cm from O, and the line OP makes an angle π/3 with the x-axis. If the electric field at P is inclined at an angle θ to the x-axis, what is θ?
• A uniformly charged semicircular wire has radius a and linear charge density λ. The magnitude of the electric field at the centre of the semicircle is
• A point Q is located on the perpendicular bisector of an electric dipole whose dipole moment is p. If Q is at a distance r from the dipole and r is much greater than the dipole size, the electric field at Q varies as

⚔️ Practice JEE Main Physics free + battle 1v1 →