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.

By Gauss's law, \(\Phi_E = \frac{q_{\text{enclosed}}}{ε_0}\). The flux depends only on the net charge enclosed inside the surface, and is independent of the size and shape of the closed surface.

The electric potential at the surface of a conducting sphere is \(V = \frac{kQ}{R} \implies kQ = VR\). The electric field at any point outside the sphere (\(r > R\)) is \(E = \frac{kQ}{r^2} = \frac{VR}{r^2}\).

For no net force, electrostatic force must equal gravitational force: \( \frac{q^2}{4 \pi \varepsilon_0 r^2} = \frac{G m^2}{r^2} \). Rearranging this formula yields \( \frac{q^2}{m^2} = 4 \pi \varepsilon_0 G \Rightarrow \frac{q}{m} = \sqrt{4 \pi \varepsilon_0 G} \).

The electrostatic force is conservative and directed radially. When moving in a circular path, the displacement is always perpendicular to the force, resulting in zero work done.

Using the work-energy theorem, \( qEd = \frac{1}{2}mv^2 \). Rearranging gives \( v = \sqrt{\frac{2qEd}{m}} = \sqrt{\frac{2 \times 3.2 \times 10^{-19} \times 1.6 \times 10^5 \times 2 \times 10^{-2}}{6.4 \times 10^{-27}}} = 4\sqrt{2} \times 10^5\text{ m/s} \).