Part (a): Electric Field due to a Uniformly Charged Infinite Plane Sheet
Step 1 (Governing Law & Gaussian Surface): [0.5 Mark]
According to Gauss's Law, the total electric flux $\phi_E$ through a closed Gaussian surface is given by:
$$\phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}$$We consider a cylindrical Gaussian surface of cross-sectional area $A$ perpendicular to the sheet, extending on both sides.
Step 2 (Mathematical Working): [1.0 Mark]
- The flux through the curved surface is zero because $\vec{E} \perp d\vec{A}$ ($\theta = 90^\circ$).
- The flux through the two flat end caps is $\phi_E = EA + EA = 2EA$.
- The charge enclosed by the cylinder is $q_{enclosed} = \sigma A$, where $\sigma$ is the surface charge density.
- Substituting into Gauss's Law:
Step 3 (Final Concluded Answer): [0.5 Mark]
Solving for $E$, we get the expression for the electric field:
$$E = \frac{\sigma}{2\epsilon_0}$$In vector form: $\vec{E} = \frac{\sigma}{2\epsilon_0} \hat{n}$, where $\hat{n}$ is a unit vector normal to the sheet.
Part (b): Two Parallel Charged Sheets
Step 1 (Principle of Superposition): [0.5 Mark]
The net electric field is the vector sum of the fields produced by individual sheets: $\vec{E}_{net} = \vec{E}_1 + \vec{E}_2$.
Step 2 (Mathematical Working): [0.5 Mark]
- (i) Inside: The fields due to the two sheets point in opposite directions. $$E_{in} = \frac{\sigma}{2\epsilon_0} - \frac{\sigma}{2\epsilon_0} = 0$$
- (ii) Outside: The fields due to the two sheets point in the same direction. $$E_{out} = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$$
Step 3 (Final Concluded Answer): [0.0 Mark - Included in Step 2]
The net electric field inside is zero, and outside is $\frac{\sigma}{\epsilon_0}$.
💡 EL10 Concept Insight: Imagine the sheets are like two pushy crowds; inside, they push against each other equally so you stay still (zero field), but outside, they both push in the same direction, making the force twice as strong!