Question

A cubical Gaussian surface has side of length a = 10 cm. Electric field lines are parallel to x-axis as shown. The magnitudes of electric fields through surfaces ABCD and EFGH are 6kNC^-1 and 9kNC^-1 respectively. Then the total charge enclosed by the cube is
[Take ε₀ = 9 × 10^-12 Fm^-1]

• A. 1.35 nC
• B. -1.35 nC
• C. 0.27 nC
• D. -0.27 nC

Answer 1

The Correct Option is C

Given data:

• Side length of the cube, \( a = 10 \) cm \( = 0.1 \) m 
• Magnitude of electric field through surface ABCD, \( E_{ABCD} = 6 \times 10^3 \) N/C
• Magnitude of electric field through surface EFGH, \( E_{EFGH} = 9 \times 10^3 \) N/C
• Permittivity of free space, \( \epsilon_0 = 9 \times 10^{-12} \) F/m

Using Gauss's Law, the total electric flux through the Gaussian surface is equal to the total charge enclosed divided by the permittivity of free space:

\(\Phi = \frac{Q_{\text{enc}}}{\epsilon_0}\)

The electric flux through a surface is given by:

\(\Phi = E \cdot A \cdot \cos(\theta)\)

Since the electric field lines are parallel to the x-axis, the angle \( \theta \) between the electric field and the normal to the surfaces ABCD and EFGH is \( 0^\circ \) and \( 180^\circ \) respectively. Therefore, \( \cos(0^\circ) = 1 \) and \( \cos(180^\circ) = -1 \).

The area of each face of the cube is:

\(A = a^2 = (0.1 \, \text{m})^2 = 0.01 \, \text{m}^2\)

The electric flux through surface ABCD is:

\(\Phi_{ABCD} = E_{ABCD} \cdot A \cdot \cos(0^\circ) = 6 \times 10^3 \, \text{N/C} \cdot 0.01 \, \text{m}^2 \cdot 1 = 60 \, \text{N} \cdot \text{m}^2/\text{C}\)

The electric flux through surface EFGH is:

\(\Phi_{EFGH} = E_{EFGH} \cdot A \cdot \cos(180^\circ) = 9 \times 10^3 \, \text{N/C} \cdot 0.01 \, \text{m}^2 \cdot (-1) = -90 \, \text{N} \cdot \text{m}^2/\text{C}\)

The total electric flux through the Gaussian surface is the sum of the fluxes through all surfaces. Since the electric field lines are parallel to the x-axis, the flux through the other four faces is zero. Therefore, the total flux is:

\(\Phi_{\text{total}} = \Phi_{ABCD} + \Phi_{EFGH} = 60 \, \text{N} \cdot \text{m}^2/\text{C} + (-90 \, \text{N} \cdot \text{m}^2/\text{C}) = -30 \, \text{N} \cdot \text{m}^2/\text{C}\)

Using Gauss's Law, the total charge enclosed by the cube is:

\(Q_{\text{enc}} = \Phi_{\text{total}} \cdot \epsilon_0 = -30 \, \text{N} \cdot \text{m}^2/\text{C} \cdot 9 \times 10^{-12} \, \text{F/m}\)

\(Q_{\text{enc}} = -270 \times 10^{-12} \, \text{C} = -0.27 \times 10^{-9} \, \text{C} = -0.27 \, \text{nC}\)

Therefore, the total charge enclosed by the cube is \( \boxed{-0.27 \, \text{nC}} \).