The electric field at a point due to a point charge is `20 NC^(-1)` and electric potential at that point is `10 JC^(-1)`. Calculate the distance of the point from the charge and the magnitude of the charge.

Explanation

To solve the problem, we will use the formulas for electric field (E) and electric potential (V) due to a point charge (Q) at a distance (r) from the charge.

### Step 1: Write down the formulas for electric field and potential.
The electric field (E) due to a point charge is given by:
\[ E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r^2} \]
The electric potential (V) due to a point charge is given by:
\[ V = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r} \]

### Step 2: Substitute the given values into the equations.
We know from the problem:
- \( E = 20 \, \text{N/C} \)
- \( V = 10 \, \text{J/C} \)

### Step 3: Rearrange the equations to find the distance (r).
From the electric field equation, we can express \( Q \) in terms of \( r \):
\[ Q = E \cdot r^2 \cdot 4\pi\epsilon_0 \]
From the potential equation, we can express \( Q \) as:
\[ Q = V \cdot r \cdot 4\pi\epsilon_0 \]

### Step 4: Set the two expressions for Q equal to each other.
Since both expressions equal \( Q \), we can set them equal:
\[ E \cdot r^2 \cdot 4\pi\epsilon_0 = V \cdot r \cdot 4\pi\epsilon_0 \]

### Step 5: Cancel \( 4\pi\epsilon_0 \) from both sides.
This gives us:
\[ E \cdot r^2 = V \cdot r \]

### Step 6: Rearrange to find r.
Dividing both sides by \( r \) (assuming \( r \neq 0 \)):
\[ E \cdot r = V \]
Now, solving for \( r \):
\[ r = \frac{V}{E} \]
Substituting the values:
\[ r = \frac{10 \, \text{J/C}}{20 \, \text{N/C}} = 0.5 \, \text{m} \]

### Step 7: Substitute r back to find Q.
Now, substitute \( r \) back into either equation for \( Q \). Using the electric field equation:
\[ Q = E \cdot r^2 \cdot 4\pi\epsilon_0 \]
Substituting \( E = 20 \, \text{N/C} \), \( r = 0.5 \, \text{m} \), and \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \):
\[ Q = 20 \cdot (0.5)^2 \cdot 4\pi(8.85 \times 10^{-12}) \]
Calculating \( (0.5)^2 = 0.25 \):
\[ Q = 20 \cdot 0.25 \cdot 4\pi(8.85 \times 10^{-12}) \]
\[ Q = 5 \cdot 4\pi(8.85 \times 10^{-12}) \]
Calculating \( 4\pi \approx 12.566 \):
\[ Q \approx 5 \cdot 12.566 \cdot 8.85 \times 10^{-12} \]
\[ Q \approx 5.55 \times 10^{-11} \, \text{C} \]

### Final Answers:
- Distance \( r = 0.5 \, \text{m} \)
- Charge \( Q \approx 5.55 \times 10^{-11} \, \text{C} \)