The electric potential at a point in free space due to a charge `Q` coulomb is `Q xx 10^(11)` volts. The electric field at that point is

Options

A. `4 pi epsilon_(0)Q xx 10^(22) V//m`
B. `12 pi epsilon_(0)Q xx 10^(20) V//m`
C. `4 pi epsilon_(0)Q xx 10^(20) V//m`
D. `12 pi epsilon_(0)Q xx 10^(22) V//m`

Explanation

To find the electric field at a point in free space due to a charge \( Q \) coulombs, given that the electric potential \( V \) at that point is \( Q \times 10^{11} \) volts, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Relationship Between Electric Potential and Electric Field:**
The electric potential \( V \) due to a point charge \( Q \) at a distance \( R \) is given by the formula:
\[ V = \frac{KQ}{R} \]
where \( K \) is Coulomb's constant, approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

2. **Setting Up the Equation:**
From the problem, we know:
\[ V = Q \times 10^{11} \]
Therefore, we can set the two expressions for potential equal to each other:
\[ Q \times 10^{11} = \frac{KQ}{R} \]

3. **Cancelling \( Q \) from Both Sides:**
Assuming \( Q \neq 0 \), we can divide both sides by \( Q \):
\[ 10^{11} = \frac{K}{R} \]

4. **Solving for \( R \):**
Rearranging the equation gives:
\[ R = \frac{K}{10^{11}} \]
Substituting \( K = 9 \times 10^9 \):
\[ R = \frac{9 \times 10^9}{10^{11}} = 9 \times 10^{-2} \, \text{m} = 0.09 \, \text{m} \]

5. **Finding the Electric Field \( E \):**
The electric field \( E \) due to a point charge is given by:
\[ E = \frac{KQ}{R^2} \]
Substituting \( R = 9 \times 10^{-2} \):
\[ E = \frac{KQ}{(9 \times 10^{-2})^2} \]
\[ E = \frac{9 \times 10^9 Q}{81 \times 10^{-4}} = \frac{9 \times 10^9 Q}{81 \times 10^{-4}} = \frac{9 \times 10^{13} Q}{81} \]
\[ E = \frac{Q \times 10^{13}}{9} \, \text{V/m} \]

6. **Expressing in Terms of \( \epsilon_0 \):**
We know that \( \frac{4}{\epsilon_0} = 9 \times 10^9 \), so we can express the electric field as:
\[ E = \frac{4}{\epsilon_0} \times \frac{10^{13}}{9} Q \]

### Final Answer:
The electric field at that point is:
\[ E = \frac{4 \times 10^{13} Q}{9 \epsilon_0} \, \text{V/m} \]