$\text{A positive point charge of}{10}^{-8}\text{C is kept at a distance of}\mathrm{ 20}\mathrm{cm }\text{from the center of a neutral conducting sphere}$

$\text{of radius}\mathrm{ 10}\mathrm{cm}.\text{The sphere is then grounded and the charge on the sphere is measured. The grounding}$

$\text{is then removed and subsequently the point charge is moved by a distance of}10\mathrm{cm}\text{further away from the center}$

$\text{of the sphere along the radial direction.}\text{ Taking }\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9\frac{N{m}^2}{C^2}\text{(where}{\epsilon }_0\text{is the permittivity of free space),}$

$\text{which of the following statements is/are correct:}$

`. Let's carefully format ``: "Before the grounding, the electrostatic potential of the sphere is 450 V ." (or all three options if provided). Let's write down the solution structure neatly adhering to all rules. Before the grounding, the electrostatic potential of the sphere is 450 V .

Correct Option/Answer:
The correct statement is: Before the grounding, the electrostatic potential of the sphere is 450 V .

Step-by-step Explanation:

1. Given Data:
- Magnitude of the positive point charge, $q={10}^{-8}\text{ C}$
- Initial distance of the point charge from the center of the sphere, $d=20\text{ cm}=0.2\text{ m}$
- Radius of the neutral conducting sphere, $R=10\text{ cm}=0.1\text{ m}$
- Electrostatic constant, $k=\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9\text{ N }{\text{m}}^2/{\text{C}}^2$

2. Electrostatic Potential of the Sphere Before Grounding:

Since the sphere is a conductor, its surface as well as its entire volume forms an equipotential region. Therefore, the potential at any point on or inside the sphere is equal to the potential at its center.

Before grounding, the sphere is uncharged (net charge on the sphere, $Q_{\text{sphere}}=0$). The potential at the center of the sphere due to its own charge is zero.

The total potential at the center of the sphere is solely due to the external point charge $q$ located at a distance $d=0.2\text{ m}$:

$V=\frac{k·q}{d}$

Substituting the given numerical values:

$V=\frac{(9\times {10}^9)\times \left({10}^{-8}\right)}{0.2}$

$V=\frac{90}{0.2}=450\text{ V}$

Thus, before grounding, the electrostatic potential of the conducting sphere is indeed 450 V.