Answer:
The peak value of the electric field is 489.64 V/m
Explanation:
Given;
power of the laser, P = 1.0 mW = 1 x 10⁻³ W
Radius of the beam, R = 1.0 mm = 1 x 10⁻³ m
Area of the beam = πr² = π(1 x 10⁻³ )² = 3.142 x 10⁻⁶ m²
The average intensity of the light = P / A
The average intensity of the light = ( 1 x 10⁻³) / (3.142 x 10⁻⁶)
The average intensity of the light = 318.27 W/m²
The peak value of the electric field is given by;
Therefore, the peak value of the electric field is 489.64 V/m.
Answer:
The correct option is (a).
Explanation:
We know that, the E is inversely proportional to the distance as follows :
We can write it as follows :
Put all the values,
So, the correct option is (a).
Answer:
The electric flux is
Explanation:
Given:
- Radius of the disc R=0.50 m
- Angle made by disk with the horizontal
- Magnitude of the electric Field
The flux of the Electric Field E due to the are dA in space can be found out by using Gauss Law which is as follows
where
- is the total Electric Flux
- E is the Electric Field
- dA is the Area through which the electric flux is to be calculated.
Now according to question we have
Hence the electric flux is calculated.