Show that average value of radiant flux density `S` over a single period 'T' is given by `S=1/(2cmu_0)E_0^2`.

Explanation

Radiant flux density `.S.` , also known as poynting vector , is given by
` vecS = ( 1) /( mu_0) ( vecE xx vecB)`
where ` vecE ` and ` vecB` are electric field vector and magnetic field , respectively .
Also ` (1)/( mu_0 ) = c^2 epsi_0 " " [ :. c= (1)/( sqrt(mu _0 epis_0))]`
` therefore vecS =- c^2 epsi_0 ( vecE xx vecB )`
Let us consider that the electromagnetic wave is propargating along the X - axis and the electric field vector ` vecE` is along the Y - axis and the electric field vector ` vecB ` is along the Z- axis
` therefore vecE = ( E_0 cos ( kx - omega t) hatj `
` vecB = ( B_0 cos ( kx - omega t) hatk `
` implies vecE xx vecB = ( E_0 B_0 ) cos ^2 ( kx - omega t) ( hatj xx hatk )`
` therefore vecS = c^2 epsi_0 ( E_0 B_0 ) cos ^2 ( kx - omega t) (hati ) `
Over the complete cycle ( time interval T) . the average value of the magnitude of radiant flux density is given by
`S_(av ) = c^2 epsi_0 ( E_0 B_0 ) (1)/(T ) int_(0) ^(T ) cos ^2 ( kx -omega t ) dt `
since ` int_(0)^(T ) cos ^2 ( kx - omega t) dt - (T)/(2 ) `
` therefore S_(Av ) = c^2 epsi_0 E_0 xx B_0 xx (1)/(T ) xx (T)/(2)`
`implies S_(Av ) = ( c^2 epsi_0 E_0 B_0 )/( 2 ) `
` = (c^2 epsi_0 E_0 ) /( 2) xx (E_0 )/(c ) [ :. B_0 = (E_(0 ))/(C)] = ( c epsi_0 E_(0)^(2))/(2)`
`=1/2 xx (c ) /( c^2 mu _0 ) E_(0)^(2) [ epsi_0 = (1)/( c^2 mu _0 ) ]`
` implies S_(av ) = (E_0^2)/( 2 mu _ 0 C)`