Solve for the electric field intensity between parallel plates, the factors affecting electric flux through a closed surface, and the effect of a dielectric on a capacitor.

where $V$ is the potential difference and $d$ is the separation.

The correct option is $\boxed{\text{400 N/C}}$.

Q.2:
The electric flux through a closed surface, according to Gauss's Law ($\Phi_E = \frac{Q_{enc}}{\epsilon_0}$), depends only on the net charge enclosed by the surface and the permittivity of the medium. It does not depend on the shape or size of the surface.
The correct option is $\boxed{\text{Shape and size of the surface}}$.

Q.3:
When a dielectric material is inserted between the plates of a charged capacitor, it becomes polarized. This polarization creates an induced electric field within the dielectric that opposes the original electric field, thereby reducing the net electric field in the region between the plates.
The correct option is $\boxed{\text{Decreases}}$.

Q.4:

The volume $V = AL$.

The resistance is quadrupled.
The correct option is $\boxed{\text{is quadrupled}}$.

Q.5:
A thermocouple generates a voltage when there is a temperature difference between two dissimilar electrical conductors. This phenomenon is known as the Seebeck effect.
The correct option is $\boxed{\text{Seebeck effect}}$.

Q.6:

where $I$ is current, $L$ is length, $B$ is magnetic field strength, and $\theta$ is the angle between the current direction and the magnetic field.

$I = 5 \text{ A}$, $L = 100 \text{ cm} = 1 \text{ m}$, $B = 4 \text{ T}$, $\theta = 45^\circ$.

The correct option is $\boxed{\text{14.14 N}}$.

Q.7:
The iron core in a transformer is laminated (made of thin, insulated sheets) to increase the electrical resistance to eddy currents. This significantly reduces the magnitude of these circulating currents, thereby minimizing the energy loss as heat due to the heating effect of eddy currents.
The correct option is $\boxed{\text{Reduce heating effect of eddy currents}}$.

Q.8:
In an RLC series AC circuit, resonance occurs when the inductive reactance ($X_L$) equals the capacitive reactance ($X_C$). The impedance $Z$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$. At resonance, $X_L - X_C = 0$, so the impedance becomes $Z = \sqrt{