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.
An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :

A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).