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.
A Wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances \[ (R_1=R_2=R_3=R_4). \] When \(R_3\) resistance is heated, its resistance value increases by \(10%\). The potential difference \((V_a-V_b)\) after \(R_3\) is heated is _______ V. 
The heat generated in 1 minute between points A and B in the given circuit, when a battery of 9 V with internal resistance of 1 \(\Omega\) is connected across these points is ______ J. 
The following diagram shows a Zener diode as a voltage regulator. The Zener diode is rated at \(V_z = 5\) V and the desired current in load is 5 mA. The unregulated voltage source can supply up to 25 V. Considering the Zener diode can withstand four times of the load current, the value of resistor \(R_s\) (shown in circuit) should be_______ \(\Omega\).
An object is projected with kinetic energy K from point A at an angle 60° with the horizontal. The ratio of the difference in kinetic energies at points B and C to that at point A (see figure), in the absence of air friction is : 