A lamp radiates power `P_(0)` uniformly in all directions, the amplitude of elctric field strength `E_(0)` at a distance `r` from it is

Options

A. `E_(0)=(P_(0))/(2piepsilon_(0)cr^(2))`
B. `E_(0)=sqrt((P_(0))/(2piepsilon_(0)cr^(2)))`
C. `E_(0)=sqrt((P_(0))/(4piepsilon_(0)cr^(2)))`
D. `E_(0)=sqrt((P_(0))/(8piepsilon_(0)cr^(2)))`

Explanation

To find the amplitude of the electric field strength \( E_0 \) at a distance \( r \) from a lamp radiating power \( P_0 \) uniformly in all directions, we can follow these steps:

### Step 1: Understand the relationship between power, intensity, and area
The intensity \( I \) of the radiation is defined as the power per unit area. For a point source radiating uniformly in all directions, the area \( A \) over which the power is spread at a distance \( r \) is the surface area of a sphere, given by:
\[ A = 4\pi r^2 \]
Thus, the intensity \( I \) can be expressed as:
\[ I = \frac{P_0}{A} = \frac{P_0}{4\pi r^2} \]

### Step 2: Relate intensity to electric field strength
The intensity \( I \) is also related to the electric field strength \( E_0 \) by the formula:
\[ I = \frac{1}{2} \epsilon_0 c E_0^2 \]
where \( \epsilon_0 \) is the permittivity of free space and \( c \) is the speed of light in vacuum.

### Step 3: Equate the two expressions for intensity
From the two expressions for intensity, we can set them equal to each other:
\[ \frac{P_0}{4\pi r^2} = \frac{1}{2} \epsilon_0 c E_0^2 \]

### Step 4: Solve for \( E_0 \)
Rearranging the equation to solve for \( E_0^2 \):
\[ E_0^2 = \frac{P_0}{2\pi r^2 \epsilon_0 c} \]
Taking the square root gives:
\[ E_0 = \sqrt{\frac{P_0}{2\pi r^2 \epsilon_0 c}} \]

### Final Result
Thus, the amplitude of the electric field strength \( E_0 \) at a distance \( r \) from the lamp is:
\[ E_0 = \sqrt{\frac{P_0}{2\pi r^2 \epsilon_0 c}} \]

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