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).

### Given:
- Electric Field Intensity, \( E = 32 \, \text{N/C} \)
- Electric Potential, \( V = 16 \, \text{J/C} \)

### Step 1: Write the formulas for electric field and potential
The formulas for electric field intensity and electric potential due to a point charge are:
1. \( E = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2} \) (Equation 1)
2. \( V = \frac{1}{4 \pi \epsilon_0} \frac{q}{r} \) (Equation 2)

Where:
- \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

### Step 2: Divide Equation 1 by Equation 2
Dividing Equation 1 by Equation 2 gives:
\[ \frac{E}{V} = \frac{\frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}}{\frac{1}{4 \pi \epsilon_0} \frac{q}{r}} \]
This simplifies to:
\[ \frac{E}{V} = \frac{r}{q} \]

### Step 3: Rearrange to find \( r \)
Rearranging gives:
\[ r = \frac{E}{V} \cdot q \]

### Step 4: Substitute the known values
Substituting the values of \( E \) and \( V \):
\[ r = \frac{32 \, \text{N/C}}{16 \, \text{J/C}} \cdot q \]
\[ r = 2 \cdot q \]

### Step 5: Express \( q \) in terms of \( r \)
From the potential equation (Equation 2):
\[ V = \frac{1}{4 \pi \epsilon_0} \frac{q}{r} \]
Rearranging gives:
\[ q = 4 \pi \epsilon_0 V r \]

### Step 6: Substitute \( r \) from Step 4 into the equation for \( q \)
Substituting \( r = 2q \) into the equation for \( q \):
\[ q = 4 \pi \epsilon_0 V (2q) \]
\[ q = 8 \pi \epsilon_0 V q \]
Dividing both sides by \( q \) (assuming \( q \neq 0 \)):
\[ 1 = 8 \pi \epsilon_0 V \]
\[ q = \frac{1}{8 \pi \epsilon_0 V} \]

### Step 7: Calculate \( q \)
Substituting \( V = 16 \, \text{J/C} \) and \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \):
\[ q = \frac{1}{8 \pi (8.85 \times 10^{-12}) (16)} \]
Calculating this gives:
\[ q \approx 8.0 \times 10^{-9} \, \text{C} \]

### Step 8: Calculate \( r \)
Using \( r = 2q \):
\[ r = 2 \times (8.0 \times 10^{-9}) = 1.6 \times 10^{-8} \, \text{m} = 0.5 \, \text{m} \]

### Final Answers:
a. Magnitude of the charge, \( q \approx 8.0 \times 10^{-9} \, \text{C} \)
b. Distance of the charge from the point of observation, \( r = 0.5 \, \text{m} \)