The electric potential on the surface of a sphere of radius R due to a charge `3 xx 10^(-6)C` is 500V. The intensity of electric field on the surface of the sphere is `(NC^(-1))` is

Options

A. `lt 10`
B. `gt 20`
C. between 10 and 20
D. `lt 5`

Explanation

To solve the problem, we need to find the electric field intensity on the surface of a sphere given the electric potential and the charge. Let's go through the steps one by one.

### Step-by-Step Solution:

**Step 1: Understand the relationship between electric potential (V) and electric field (E).**
The electric potential \( V \) at the surface of a sphere due to a point charge is given by the formula:
\[ V = \frac{1}{4 \pi \epsilon_0} \frac{Q}{R} \]
where \( Q \) is the charge, \( R \) is the radius of the sphere, and \( \epsilon_0 \) is the permittivity of free space.

**Step 2: Use the given values to set up the equation.**
From the problem, we know:
- \( Q = 3 \times 10^{-6} \, C \)
- \( V = 500 \, V \)

Substituting these values into the potential formula:
\[ 500 = \frac{1}{4 \pi \epsilon_0} \frac{3 \times 10^{-6}}{R} \]

**Step 3: Rearranging the equation to find R.**
To find \( R \), we can rearrange the equation:
\[ R = \frac{1}{4 \pi \epsilon_0} \frac{3 \times 10^{-6}}{500} \]

**Step 4: Calculate \( R \).**
Using the value of \( \epsilon_0 \approx 8.85 \times 10^{-12} \, C^2/(N \cdot m^2) \):
\[ R = \frac{(3 \times 10^{-6})}{500 \times (4 \pi \times 8.85 \times 10^{-12})} \]
Calculating the denominator:
\[ 4 \pi \epsilon_0 \approx 4 \times 3.14 \times 8.85 \times 10^{-12} \approx 1.11 \times 10^{-10} \]
Now substituting this back:
\[ R = \frac{3 \times 10^{-6}}{500 \times 1.11 \times 10^{-10}} \approx \frac{3 \times 10^{-6}}{5.55 \times 10^{-8}} \approx 54.05 \, m \]

**Step 5: Calculate the electric field intensity (E) on the surface of the sphere.**
The electric field \( E \) at the surface of the sphere is given by:
\[ E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{R^2} \]
Substituting the values we have:
\[ E = \frac{1}{4 \pi \epsilon_0} \frac{3 \times 10^{-6}}{(54.05)^2} \]
Calculating \( R^2 \):
\[ R^2 \approx (54.05)^2 \approx 2927.60 \]
Now substituting:
\[ E \approx \frac{(3 \times 10^{-6})}{(4 \pi \times 8.85 \times 10^{-12}) \times 2927.60} \]
Calculating the denominator:
\[ E \approx \frac{3 \times 10^{-6}}{1.11 \times 10^{-10} \times 2927.60} \approx \frac{3 \times 10^{-6}}{3.25 \times 10^{-7}} \approx 9.23 \times 10^{1} \, N/C \]

### Final Answer:
The intensity of the electric field on the surface of the sphere is approximately \( 92.3 \, N/C \).