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 \( \boxed{-0.27 \, \text{nC}} \).

Answer 2

We can use Gauss's Law to find the total charge enclosed by the cube. Gauss's law states that: \[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \] where:
\(\vec{E}\) is the electric field,
\(d\vec{A}\) is the differential area vector,
\(Q_{\text{enc}}\) is the total charge enclosed,
\(\epsilon_0 = 9 \times 10^{-12} \, \text{Fm}^{-1}\) is the permittivity of free space.

Since the electric field lines are parallel to the x-axis, the flux through the faces of the cube is calculated by the product of the electric field and the area of the faces through which the field lines pass. The electric field through faces ABCD and EFGH are given as: - \(E_{\text{ABCD}} = 6 \, \text{kN/C}\) - \(E_{\text{EFGH}} = 9 \, \text{kN/C}\) The area of each face of the cube is: \[ A = a^2 = (0.1 \, \text{m})^2 = 0.01 \, \text{m}^2 \] The flux through the faces ABCD and EFGH is: \[ \Phi_{\text{ABCD}} = E_{\text{ABCD}} \times A = 6 \times 10^3 \, \text{N/C} \times 0.01 \, \text{m}^2 = 60 \, \text{Nm}^2/\text{C} \] \[ \Phi_{\text{EFGH}} = E_{\text{EFGH}} \times A = 9 \times 10^3 \, \text{N/C} \times 0.01 \, \text{m}^2 = 90 \, \text{Nm}^2/\text{C} \] Now, applying Gauss's law: \[ \Phi_{\text{total}} = \Phi_{\text{ABCD}} + \Phi_{\text{EFGH}} = 60 + 90 = 150 \, \text{Nm}^2/\text{C} \] Using Gauss's law: \[ Q_{\text{enc}} = \Phi_{\text{total}} \times \epsilon_0 = 150 \, \text{Nm}^2/\text{C} \times 9 \times 10^{-12} \, \text{Fm}^{-1} \] \[ Q_{\text{enc}} = 1.35 \times 10^{-12} \, \text{C} = 0.27 \, \text{nC} \] Thus, the correct answer is: \[{\text{(C) } 0.27 \, \text{nC}} \]