Step 1: Consider a uniform surface charge density \( \sigma \) on both a conducting plate and a nonconducting sheet.
For a nonconducting sheet, the electric field at a point near the surface can be derived from Gauss's law.
Using a Gaussian pillbox with a small area \( A \) around the surface, we apply Gauss’s law:
Since the sheet is nonconducting, the charge is only on one side of the sheet.
The total enclosed charge is \( \sigma A \).
The electric flux through the pillbox is:
\[ E \cdot A + E \cdot A = \frac{\sigma A}{\epsilon_0} \] \[ 2E = \frac{\sigma}{\epsilon_0} \quad \Rightarrow \quad E = \frac{\sigma}{2 \epsilon_0} \] So, the electric field due to a nonconducting sheet with charge density \( \sigma \) is:
\[ E_{non-conducting}} = \frac{\sigma}{2 \epsilon_0} \] For a conducting plate, the situation is different because charges on a conductor move freely and spread out evenly. The field due to a conducting plate is calculated similarly using Gauss’s law. For a conducting plate, the charge distributes evenly on both sides of the plate, so the electric field is the sum of the fields from both sides. Each side contributes \( \frac{\sigma}{2 \epsilon_0} \), and thus the total electric field is:
\[ E_{conducting}} = \frac{\sigma}{\epsilon_0} \]
Step 2: We can now compare the electric fields for both cases:
For a nonconducting sheet, \( E_{non-conducting}} = \frac{\sigma}{2 \epsilon_0} \).
For a conducting plate, \( E_{conducting}} = \frac{\sigma}{\epsilon_0} \).
Therefore, the electric field for a conducting plate is twice that for a nonconducting sheet. \[ E_{conducting}} = 2 \cdot E_{non-conducting}} \] Thus, the electric field is twice in the case of a conducting plate compared to a nonconducting sheet.
Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R):
Assertion (A): In an insulated container, a gas is adiabatically shrunk to half of its initial volume. The temperature of the gas decreases.
Reason (R): Free expansion of an ideal gas is an irreversible and an adiabatic process. \text{In the light of the above statements, choose the correct answer from the options given below:}
For the circuit shown above, the equivalent gate is:
Find the equivalent resistance between two ends of the following circuit:
The circuit consists of three resistors, two of \(\frac{r}{3}\) in series connected in parallel with another resistor of \(r\).