At s point due to a point charge, the values of electric field intensity and potential are `32 NC^-1` and `16 JC^-1`, respectively. Calculate the
a. magnitude of the charge, and
b. distance of the charge from the point of observation.

Explanation

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

### Given:
- Electric field intensity, \( E = 32 \, \text{N/C} \)
- Electric potential, \( V = 16 \, \text{J/C} \)

### Formulas:
1. The formula for electric field intensity due to a point charge:
\[ E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{R^2} \tag{1} \]
2. The formula for electric potential due to a point charge:
\[ V = \frac{1}{4 \pi \epsilon_0} \frac{Q}{R} \tag{2} \]

### Step 1: Divide Equation (1) by Equation (2)
To eliminate \( Q \) and find \( R \), we can divide the equation for electric field by the equation for electric potential:

\[ \frac{E}{V} = \frac{\frac{1}{4 \pi \epsilon_0} \frac{Q}{R^2}}{\frac{1}{4 \pi \epsilon_0} \frac{Q}{R}} \]

This simplifies to:

\[ \frac{E}{V} = \frac{R}{Q} \]

Rearranging gives:

\[ R = \frac{E}{V} \cdot Q \tag{3} \]

### Step 2: Substitute the values of E and V
Substituting the values of \( E \) and \( V \):

\[ R = \frac{32}{16} \cdot Q \]

This simplifies to:

\[ R = 2Q \tag{4} \]

### Step 3: Substitute R back into Equation (1) or (2) to find Q
Now, we can substitute \( R \) back into either equation (1) or (2) to find \( Q \). We will use equation (1):

\[ 32 = \frac{1}{4 \pi \epsilon_0} \frac{Q}{R^2} \]

Substituting \( R = 2Q \):

\[ 32 = \frac{1}{4 \pi \epsilon_0} \frac{Q}{(2Q)^2} \]

This simplifies to:

\[ 32 = \frac{1}{4 \pi \epsilon_0} \frac{Q}{4Q^2} \]

\[ 32 = \frac{1}{16 \pi \epsilon_0 Q} \]

### Step 4: Solve for Q
Rearranging gives:

\[ Q = \frac{1}{16 \pi \epsilon_0 \cdot 32} \]

Using \( \epsilon_0 \approx 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \):

\[ Q = \frac{1}{16 \cdot \pi \cdot 8.85 \times 10^{-12} \cdot 32} \]

Calculating this gives:

\[ Q \approx 8.85 \times 10^{-9} \, \text{C} \]

### Step 5: Find R using the value of Q
Now substitute \( Q \) back into equation (4) to find \( R \):

\[ R = 2 \cdot 8.85 \times 10^{-9} \approx 1.77 \times 10^{-8} \, \text{m} \]

### Final Answers:
a. Magnitude of the charge, \( Q \approx 8.85 \times 10^{-9} \, \text{C} \)

b. Distance of the charge from the point of observation, \( R \approx 1.77 \times 10^{-8} \, \text{m} \)