The energy density of an EM wave is proportional to the square of the electric f ield. Squaring a sinusoidal function doubles the frequency.
The energy density of the wave is given by:
\( \text{Energy density} = \frac{1}{2} \varepsilon_0 E_{\text{net}}^2 \)
Substitute \( E_{\text{net}} = E_0 \sin(\omega t - kx) \):
\[ \text{Energy density} = \frac{1}{2} \varepsilon_0 E_0^2 \sin^2(\omega t - kx) \]
Using the trigonometric identity \( \sin^2 x = \frac{1}{2}(1 - \cos 2x) \):
\[ \sin^2(\omega t - kx) = \frac{1}{2}(1 - \cos(2\omega t - 2kx)) \]
Substitute this into the energy density formula:
\[ \text{Energy density} = \frac{1}{2} \varepsilon_0 E_0^2 \cdot \frac{1}{2}(1 - \cos(2\omega t - 2kx)) \]
Simplify the expression:
\[ \text{Energy density} = \frac{1}{4} \varepsilon_0 E_0^2 (1 - \cos(2\omega t - 2kx)) \]
The energy density of the wave is:
\( \frac{1}{4} \varepsilon_0 E_0^2 (1 - \cos(2\omega t - 2kx)) \)
Using a variable frequency ac voltage source the maximum current measured in the given LCR circuit is 50 mA for V = 5 sin (100t) The values of L and R are shown in the figure. The capacitance of the capacitor (C) used is_______ µF.


Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
