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. `12piepsi_(0)Qxx10^(22)Vm^(-1)`
B. `4 pi epsi_(0)Qxx10^(22)Vm^(-1)`
C. `12pi epsi_(0)Qxx10^(20)V m^(-1)`
D. `4 pi epsi _(0)Qxx10^(20)V m^(-1)`

Explanation

To find the electric field at a point in free space due to a charge \( Q \) coulombs when the electric potential \( V \) at that point is given as \( V = Q \times 10^{11} \) volts, we can use the relationships between electric potential and electric field.

### Step-by-Step Solution:

1. **Understanding Electric Potential and Electric Field**:
The electric potential \( V \) at a distance \( R \) from a point charge \( Q \) is given by the formula:
\[ V = \frac{KQ}{R} \]
where \( K \) is Coulomb's constant, \( K = \frac{1}{4\pi \epsilon_0} \).

2. **Electric Field Formula**:
The electric field \( E \) due to a point charge \( Q \) at a distance \( R \) is given by:
\[ E = \frac{KQ}{R^2} \]

3. **Relating \( R \) to \( V \)**:
From the potential formula, we can express \( R \) in terms of \( V \):
\[ R = \frac{KQ}{V} \]

4. **Substituting \( R \) in the Electric Field Formula**:
Now, substituting \( R \) in the electric field formula:
\[ E = \frac{KQ}{\left(\frac{KQ}{V}\right)^2} \]
Simplifying this gives:
\[ E = \frac{KQ \cdot V^2}{(KQ)^2} = \frac{V^2}{KQ} \]

5. **Substituting the Given Value of \( V \)**:
We know \( V = Q \times 10^{11} \), so substituting this into the equation:
\[ E = \frac{(Q \times 10^{11})^2}{KQ} \]
This simplifies to:
\[ E = \frac{Q^2 \times 10^{22}}{KQ} = \frac{Q \times 10^{22}}{K} \]

6. **Substituting the Value of \( K \)**:
Recall that \( K = \frac{1}{4\pi \epsilon_0} \):
\[ E = 4\pi \epsilon_0 Q \times 10^{22} \]

7. **Final Result**:
Therefore, the electric field at that point is:
\[ E = 4\pi \epsilon_0 Q \times 10^{22} \text{ volts per meter} \]