At a distance `r_(1)` from a point charge, the magnitude of the electric field created by the charge is 248 N/C. At a distance `r_(2)` from the charge, the field has a magnitude of 132 N/C. Find the ratio `r_(2)//r_(1)` ?

Options

A. `0.730`
B. `0.939`
C. `1.12`
D. `1.37`

Answer

To find the ratio \( \frac{r_2}{r_1} \) given the electric fields at two distances from a point charge, we can follow these steps:

### Step 1: Write the formula for electric field due to a point charge
The electric field \( E \) created by a point charge \( q \) at a distance \( r \) is given by the formula:
\[ E = \frac{k \cdot q}{r^2} \]
where \( k \) is Coulomb's constant.

### Step 2: Set up the equations for the two distances
For the first distance \( r_1 \), the electric field \( E_1 \) is:
\[ E_1 = \frac{k \cdot q}{r_1^2} = 248 \, \text{N/C} \]
For the second distance \( r_2 \), the electric field \( E_2 \) is:
\[ E_2 = \frac{k \cdot q}{r_2^2} = 132 \, \text{N/C} \]

### Step 3: Form a ratio of the two electric field equations
We can form a ratio of the two equations:
\[ \frac{E_1}{E_2} = \frac{\frac{k \cdot q}{r_1^2}}{\frac{k \cdot q}{r_2^2}} = \frac{r_2^2}{r_1^2} \]
Substituting the values of \( E_1 \) and \( E_2 \):
\[ \frac{248}{132} = \frac{r_2^2}{r_1^2} \]

### Step 4: Simplify the ratio
Now, simplify \( \frac{248}{132} \):
\[ \frac{248}{132} = \frac{62}{33} \]
Thus, we have:
\[ \frac{r_2^2}{r_1^2} = \frac{62}{33} \]

### Step 5: Take the square root to find the ratio of distances
To find \( \frac{r_2}{r_1} \), we take the square root of both sides:
\[ \frac{r_2}{r_1} = \sqrt{\frac{62}{33}} \]

### Final Result
Thus, the ratio \( \frac{r_2}{r_1} \) is:
\[ \frac{r_2}{r_1} = \sqrt{\frac{62}{33}} \]