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: \( \frac{q}{2\epsilon_0} \)
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 $\vec{E}=E_0 x \hat{ t }$ Where $E_0=4 \times 10^4 NC ^{-1} m ^{-1}$ If $a=2 cm$, the charge contained in the cubical volume is $Q \times 10^{-14} C$ The value of $Q$ is ___ Take \(E_{0}=9\times 10^{-2}C^{2}/Nm^{2}\)
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 | \(\oint \vec{E} \cdot d \vec{s}=\frac{q}{\epsilon_0}\) |