The electric field at the surface of a charged spherical conductor is 10 kV/m. The electric field at an outward radial distance equal to the radius from its surface will be

Options

A. 1.5 kW
B. 2.5 kV/m
C. 4kV/m
D. 5kV/m

Explanation

To solve the problem, we need to determine the electric field at a point that is at a radial distance equal to the radius of the charged spherical conductor from its surface. Let's break this down step by step.

### Step 1: Understand the Given Information
We know that the electric field at the surface of the charged spherical conductor (let's call this point A) is given as:
\[ E_A = 10 \, \text{kV/m} = 10,000 \, \text{V/m} \]

### Step 2: Define the Distances
Let the radius of the spherical conductor be \( R \). The point we are interested in (let's call this point P) is at a distance \( R \) from the surface of the sphere. Therefore, the distance from the center of the sphere to point P is:
\[ OP = R + R = 2R \]

### Step 3: Use the Formula for Electric Field
The electric field \( E \) due to a charged spherical conductor at a distance \( r \) from its center is given by:
\[ E = \frac{kQ}{r^2} \]
where \( k \) is Coulomb's constant and \( Q \) is the charge of the sphere.

### Step 4: Relate the Electric Fields at Points A and P
Using the formula for electric field, we can express the electric field at point A (at the surface) and point P (at distance \( 2R \)):
- At point A (surface):
\[ E_A = \frac{kQ}{R^2} \]
- At point P:
\[ E_P = \frac{kQ}{(2R)^2} = \frac{kQ}{4R^2} \]

### Step 5: Relate the Two Electric Fields
Now, we can relate \( E_A \) and \( E_P \):
\[ E_P = \frac{E_A}{4} \]

### Step 6: Substitute the Known Value
Substituting the value of \( E_A \):
\[ E_P = \frac{10,000 \, \text{V/m}}{4} = 2,500 \, \text{V/m} \]

### Conclusion
Thus, the electric field at the point P, which is at a radial distance equal to the radius from the surface of the charged spherical conductor, is:
\[ E_P = 2,500 \, \text{V/m} \]

### Final Answer
The electric field at an outward radial distance equal to the radius from its surface will be **2,500 V/m**.

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