For a point \( Q \) on the perpendicular bisector of an electric dipole (distance \( r \) from the center, where \( r \gg \text{dipole length} \)):

1. Electric Field on Perpendicular Bisector: The electric field \( E \) at a point on the perpendicular bisector of a dipole is given by:
\[
E \propto \frac{p}{r^3}
\]

2. Dependence:
- Directly proportional to the dipole moment \( p \).
- Inversely proportional to \( r^3 \).

Answer:
The electric field at \( Q \) is proportional to:
\[
p \quad \text{and} \quad r^{-3}
\]

The electric potential \( V \) at a point \( P \) due to an electric dipole with dipole moment \( \overrightarrow{p} \) is given by:

\[
V = \frac{k \, \overrightarrow{p} \cdot \overrightarrow{r}}{r^3}
\]

Explanation:
1. Dipole Moment: The dipole moment \( \overrightarrow{p} = q \cdot d \), where \( q \) is the charge and \( d \) is the separation between charges.

2. Position Vector \( \overrightarrow{r} \): This is the vector from the center of the dipole to the point \( P \).

3. Dot Product: The potential depends on the angle between \( \overrightarrow{p} \) and \( \overrightarrow{r} \), hence \( \overrightarrow{p} \cdot \overrightarrow{r} = p r \cos \theta \).

4. Result: The formula is:
\[
V = \frac{k (\overrightarrow{p} \cdot \overrightarrow{r})}{r^3}
\]

This matches the correct answer:
\[
V = \frac{k (\overrightarrow{p} \cdot \overrightarrow{r})}{r^3}
\]

Since the electric field \( E \) is parallel to the cylinder's axis, the field lines enter and exit symmetrically through the two flat circular ends. The curved surface has no perpendicular component of the field, so no flux passes through it.

By Gauss's law, the net flux through the entire surface is:
\[
\Phi = \text{Charge enclosed} / \varepsilon_0
\]

As no charge is enclosed, \( \Phi = 0 \).

The electric flux through the hemisphere is due to the uniform electric field passing perpendicularly through the circular plane of radius \( R \).

Flux through the circular plane:
\[
\Phi = E \cdot \text{Area of circular plane} = E \cdot \pi R^2
\]

Since no flux passes through the curved surface (field is parallel to it), the total flux is:
\[
\Phi = \pi R^2 E
\]

Gauss's law states that the total electric flux passing through a closed surface is equal to the charge enclosed within the surface divided by the permittivity of the medium.

In short: The net electric field around a closed surface depends on the charge inside it.

The question states the mathematical form of Gauss's law as:

\[
\varepsilon_{0}\oint \overrightarrow{E} \cdot d\overrightarrow{S} = q
\]

Where:
- \(\varepsilon_0\) is the permittivity of free space,
- \(\overrightarrow{E}\) is the electric field vector,
- \(d\overrightarrow{S}\) is the infinitesimal area vector on the Gaussian surface,
- \(q\) is the total charge enclosed within the Gaussian surface.

Key Points in Gauss's Law:

1. The electric flux through the Gaussian surface (\(\oint \overrightarrow{E} \cdot d\overrightarrow{S}\)) depends only on the charge enclosed (\(q\)) within the surface.

2. The electric field \(\overrightarrow{E}\) at any point on the Gaussian surface depends on all charges in the system—both inside and outside the Gaussian surface.

Why Does \(E\) Depend on Charges Outside the Gaussian Surface?

While Gauss's law calculates flux based only on enclosed charge, the electric field \(\overrightarrow{E}\) at a point on the Gaussian surface is influenced by all charges, regardless of their location (inside or outside the surface). Here's why:

- Charges inside the Gaussian surface: These contribute directly to the net flux as per Gauss's law.
- Charges outside the Gaussian surface: These do not contribute to the net flux (their contributions cancel out overall due to symmetry), but they **still influence the local value of \(\overrightarrow{E}\)**.

Example:

- Imagine a spherical Gaussian surface around a point charge \(q_1\). If another charge \(q_2\) is placed outside the sphere, it doesn't affect the total flux, but it does contribute to the electric field at various points on the sphere.

Thus, the electric field \(\overrightarrow{E}\) depends on all charges in the vicinity, while the total flux (as per Gauss's law) depends solely on the charges enclosed. This is why the correct answer is:

"E depends on the charge which is inside and outside the Gaussian surface."

Option 2 is wrong because electric field lines emerge out of +ve charge and submerge into -ve charge.

In option 3 number of electric field lines is proportional to amount of charge. From +5C charge 6 field lines are emerging out where as from 1C charge only one electric field line is present.

Number of Electric field line in space represents electric field intensity. As number of field line is maximum at A followed by B then at C. so, \(E_{A}>E_{B}>E_{C}\)