How much positive charge should be given to the earth so as to have the same potential as that of a positively charged sphere of `1 muC` and radius 1 cm . ( Radius of earth = 6400km )

Options

A. 600 C
B. 640 C
C. 340 C
D. 240 C

Answer

To solve the problem of how much positive charge should be given to the Earth so that it has the same potential as a positively charged sphere of 1 µC and radius 1 cm, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Formula for Electric Potential**:
The electric potential \( V \) due to a point charge or a charged sphere is given by the formula:
\[ V = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{R} \]
where \( Q \) is the charge and \( R \) is the radius from the center of the charge to the point where the potential is being calculated.

2. **Identifying the Charges and Radii**:
- For the charged sphere:
- Charge \( Q_1 = 1 \, \mu C = 1 \times 10^{-6} \, C \)
- Radius \( R_1 = 1 \, cm = 0.01 \, m \)
- For the Earth:
- Let the charge on the Earth be \( Q_2 \).
- Radius of the Earth \( R_2 = 6400 \, km = 6400 \times 10^3 \, m \)

3. **Setting the Potentials Equal**:
Since we want the potential of the Earth to equal that of the charged sphere, we set the potentials equal:
\[ \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q_1}{R_1} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q_2}{R_2} \]
The \( \frac{1}{4 \pi \epsilon_0} \) terms cancel out:
\[ \frac{Q_1}{R_1} = \frac{Q_2}{R_2} \]

4. **Substituting the Known Values**:
Substitute \( Q_1 \), \( R_1 \), and \( R_2 \) into the equation:
\[ \frac{1 \times 10^{-6}}{0.01} = \frac{Q_2}{6400 \times 10^3} \]

5. **Cross-Multiplying to Solve for \( Q_2 \)**:
Cross-multiplying gives:
\[ Q_2 = \frac{1 \times 10^{-6} \cdot (6400 \times 10^3)}{0.01} \]

6. **Calculating \( Q_2 \)**:
Simplifying the right side:
\[ Q_2 = \frac{1 \times 10^{-6} \cdot 6400 \times 10^3}{0.01} = 6400 \times 10^{-6} \times 10^5 = 640 \, C \]

### Final Answer:
The amount of positive charge that should be given to the Earth is:
\[ \boxed{640 \, C} \]