As shown in figure, a cuboid lies in a region with electric field $\mathrm{E}=2{\mathrm{x}}^2\widehat{\mathrm{i}}-4\mathrm{y}\widehat{\mathrm{j}}+6\hat{\mathrm{k}}\mathrm{N}/\mathrm{C}$.The magnitude of charge within the cuboid is n ${\in }_0$ C. The value of n is_______ (if dimension of cuboid is $1 \mathrm{x} 2 \mathrm{x} 3{\mathrm{m}}^3$).

Given:

• Electric field: E = 2x² î − 4y ĵ + 6 k̂ (N/C)
• Dimensions of the cuboid: 1 × 2 × 3 m³

Solution:

Step 1: Gauss's Law

According to Gauss's law, the net electric flux through the cuboid is related to the enclosed charge:

Φ_net = (q / ε₀)

Where:

• Φ_net is the net electric flux.
• q is the charge enclosed by the cuboid.
• ε₀ is the permittivity of free space.

Step 2: Electric Flux Calculation

The electric field varies across the surfaces of the cuboid. For simplicity, we compute the flux through each pair of opposite surfaces:

Surface 1 (Perpendicular to x-axis):

The electric field along the x-axis is E_x = 2x². The flux through the surfaces at x = 0 and x = 1 is:

At x = 0: E_x = 2(0)² = 0

At x = 1: E_x = 2(1)² = 2

Area of each surface: A = (2 × 3) = 6 m²

Flux through x-surfaces: Φ_x = (2 × 6) - (0 × 6) = 12 N·m²/C

Surface 2 (Perpendicular to y-axis):

The electric field along the y-axis is E_y = -4y. The flux through the surfaces at y = 0 and y = 2 is:

At y = 0: E_y = -4(0) = 0

At y = 2: E_y = -4(2) = -8

Area of each surface: A = (1 × 3) = 3 m²

Flux through y-surfaces: Φ_y = (0 × 3) - (-8 × 3) = -24 N·m²/C

Surface 3 (Perpendicular to z-axis):

The electric field along the z-axis is constant: E_z = 6. The flux through the surfaces at z = 0 and z = 3 is:

At z = 0: E_z = 6

At z = 3: E_z = 6

Area of each surface: A = (1 × 2) = 2 m²

Flux through z-surfaces: Φ_z = (6 × 2) - (6 × 2) = 0 N·m²/C

Step 3: Total Flux

The total flux through the cuboid is the sum of fluxes through all surfaces:

Φ_net = Φ_x + Φ_y + Φ_z

Φ_net = 12 - 24 + 0 = -12 N·m²/C

Step 4: Enclosed Charge

Using Gauss's law:

q = ε₀ Φ_net

Substitute Φ_net = -12:

q = -12ε₀

Final Answer:

The magnitude of the enclosed charge is:

|q| = 12ε₀ C