The electric potential at a certain distance from a point charge is 600 V and the electric field is `200 NC^(-1)`. Which of the following statements will be true?

Options

A. The work done in moving a point charge of `1muC` from the given point to a point at a distance of 9 m will be `4 xx 10^(-4)J`.
B. The distance of the given point from the charge is 3 m.
C. The potential at a distance of 9 m will be 200 V.
D. The magnitude of charge is `0.2 xx 10^(-3)`C.

Explanation

To solve the problem, we will analyze the given information about the electric potential and electric field due to a point charge and determine which statements are true based on the calculations.

### Step-by-Step Solution:

1. **Given Information:**
- Electric potential (V) at a certain distance from the point charge = 600 V
- Electric field (E) at that point = 200 N/C

2. **Formulas:**
- The electric field (E) due to a point charge (Q) at a distance (R) is given by:
\[ E = \frac{kQ}{R^2} \]
- The electric potential (V) due to a point charge (Q) at a distance (R) is given by:
\[ V = \frac{kQ}{R} \]
where \( k \) is the electrostatic constant \( (k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2) \).

3. **Setting Up the Equations:**
- From the electric field:
\[ E = \frac{kQ}{R^2} \quad \text{(1)} \]
Substituting \( E = 200 \, \text{N/C} \):
\[ 200 = \frac{kQ}{R^2} \]
- From the electric potential:
\[ V = \frac{kQ}{R} \quad \text{(2)} \]
Substituting \( V = 600 \, \text{V} \):
\[ 600 = \frac{kQ}{R} \]

4. **Dividing Equation (1) by Equation (2):**
\[ \frac{E}{V} = \frac{\frac{kQ}{R^2}}{\frac{kQ}{R}} \Rightarrow \frac{E}{V} = \frac{1}{R} \]
Substituting the values:
\[ \frac{200}{600} = \frac{1}{R} \Rightarrow \frac{1}{3} = \frac{1}{R} \]
Therefore, \( R = 3 \, \text{m} \).

5. **Finding the Magnitude of Charge (Q):**
- From equation (2):
\[ 600 = \frac{kQ}{3} \]
Rearranging gives:
\[ kQ = 600 \times 3 = 1800 \]
Thus:
\[ Q = \frac{1800}{k} = \frac{1800}{9 \times 10^9} = 0.2 \times 10^{-6} \, \text{C} = 0.2 \, \mu\text{C} \]

6. **Finding Potential at a Distance of 9 m:**
- Using the formula for potential:
\[ V_{9m} = \frac{kQ}{9} \]
Substituting the values:
\[ V_{9m} = \frac{9 \times 10^9 \times 0.2 \times 10^{-6}}{9} = 0.2 \times 10^3 = 200 \, \text{V} \]

7. **Calculating Work Done in Moving a Charge:**
- The work done (W) in moving a charge \( q \) from potential \( V_A \) to \( V_B \):
\[ W = q(V_A - V_B) \]
Given \( q = 1 \, \mu\text{C} = 1 \times 10^{-6} \, \text{C} \), \( V_A = 600 \, \text{V} \), and \( V_B = 200 \, \text{V} \):
\[ W = 1 \times 10^{-6} (600 - 200) = 1 \times 10^{-6} \times 400 = 4 \times 10^{-4} \, \text{J} \]

### Conclusion:
- The distance of the given point from the charge is **3 meters** (Option B is correct).
- The potential at a distance of **9 meters** is **200 V** (Option C is correct).
- The magnitude of the charge is **0.2 µC** (Option D is correct).
- The work done in moving a point charge of **1 µC** is **4 x 10^-4 J** (Option A is correct).