Question

Light with an energy flux of $ 25 \times 10^4 W m^{-2} $ falls on a perfectly reflecting surface at normal incidence. If the surface area is $ 15 c m^2 $, the average force exerted on the surface is

• A. $1.25 \times 10^{-6} N $
• B. $2.50\times10^{-6} N$
• C. $1.20\times10^{-6}N$
• D. $3.0\times10^{-6}N$

Answer 1

The Correct Option is B

Here,
Energy flux, $I= 25 \times {10}^4 W m^{-2}$
Area, $ A = 15 c m^2 = 15 \times {10} ^{-4} m^2 $
Speed of light, $ c= 3 \times {10}^8 m s^{-1}$
For a perfectly reflecting surface, the average force exerted on the surface is
$ F= \frac{ 2IA}{ c} = \frac{ 2 \times 25 \times 10^{4} W m^{-2} \times 15 \times 10^{-4} m^{2} }{ 3 \times 10^{8} \,m\, s^{-1}}$
$ = 250 \times {10}^{-8} N = 2.50 \times {10}^{-6} \,N $

Answer 2

Ans. Electric Flux is the rate of flow of an electric field through an area. Electric flux is proportional to the number of electric field lines passing through a virtual surface. 

• Flux refers to the presence of a force field in a physical medium.
• In electronics, flux refers to an electrostatic field and any magnetic field.
• Flux is shown as "lines" in a plane that contains or intersects electric charge poles or magnetic poles.
• The total number of electric field lines passing a given area in a unit of time is defined as the electric flux.

Similar to the example above, if the plane is normal to the flow of the electric field, the total flux is given as ɸ_P = EA

The S.I. unit of electric flux is Volt Metres (V m). The dimensions of the electric flux physics are  NC⁻¹ m² or Kg m³ s⁻³ A⁻¹.

In Electrostatics, Electric Flux is described as several electric field lines, passing per unit area. It is an additional physical quantity that is used to measure the strength of an electric field and build the fundamentals of electrostatics in the world of Physics. Electric Flux can also be described as the product of the electric field and surface area projected in a direction perpendicular to the electric field.

On tilting this plane at an angle of θ, the calculated area is given by Acosθ and the total flux over this surface is given as:

ɸ_P = EACosΘ

Where,

• E = magnitude of the electric field
• A = area of the surface through which the electric flux is to be determined
• Θ = angle made by the plane and the axis parallel to the direction of flow of the electric field