1. **Using Gauss’s Law:**
When a charge \( q \) is placed at the center of one face of a cube, it can be visualized that the charge \( q \) contributes equally to two adjacent cubes.
2. **Flux Calculation:**
According to Gauss’s law, the total flux \( \Phi \) due to charge \( q \) in a closed surface is given by:
\[ \Phi_{\text{total}} = \frac{q}{\epsilon_0}. \] Since the charge \( q \) is shared equally between two adjacent cubes, the flux through each cube is:
\[ \Phi = \frac{q}{2\epsilon_0}. \]
Answer: \( \frac{q}{2\epsilon_0} \)
LIST I | LIST II | ||
A | Gauss's Law in Electrostatics | I | \(\oint \vec{E} \cdot d \vec{l}=-\frac{d \phi_B}{d t}\) |
B | Faraday's Law | II | \(\oint \vec{B} \cdot d \vec{A}=0\) |
C | Gauss's Law in Magnetism | III | \(\oint \vec{B} \cdot d \vec{l}=\mu_0 i_c+\mu_0 \in_0 \frac{d \phi_E}{d t}\) |
D | Ampere-Maxwell Law | IV | \(\oint \vec{E} \cdot d \vec{s}=\frac{q}{\epsilon_0}\) |
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: \[ \vec{E} = E_0 x \hat{i} \]
Where \( E_0 = 4 \times 10^4 \, \text{NC}^{-1} \, \text{m}^{-1} \). If \( a = 2 \, \text{cm} \), the charge contained in the cubical volume is \( Q \times 10^{-14} \, \text{C} \). The value of \( Q \) is ______. (Take \( \epsilon_0 = 9 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 \))
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).