You visited us 4 times! Enjoying our articles? Unlock Full Access!

Maxwell's Equations

Which of the ...

Question

Which of the following relations is correct for electromagnetic waves:

\frac{\partial^{2} E}{\partial x^{2}} = \frac{1}{c^{2}} \frac{\partial^{2} E}{\partial t^{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{\partial^{2} E}{\partial x^{2}} = {(\frac{\partial E}{\partial t})}^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{\partial^{2} E}{\partial x^{2}} = \frac{c^{2} \partial^{2} E}{\partial t^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{\partial^{2} E}{\partial x^{2}} = c \frac{\partial^{2} E}{\partial t^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{\partial^{2} E}{\partial x^{2}} = \frac{1}{c^{2}} \frac{\partial^{2} E}{\partial t^{2}}$
The equation of electromagnetic wave propagating along x direction in vaccum is: $\frac{\partial^{2} E}{\partial x^{2}} = \frac{1}{c^{2}} \frac{\partial^{2} E}{\partial t^{2}}$

Suggest Corrections

Similar questions

x {log}_{e} ({log}_{e} x) - x^{2} + y^{2} = 4 (y > 0)

\frac{d y}{d x}

x = e

Given :

Oxidation :

$H_{2} O_{2} \to O_{2} + 2 H^{\oplus} + 2 e^{-}$

$E^{⊖} = - 0.69 V$ ;

$2 F^{⊖} \to F_{2} + 2 e^{-}$

$E^{⊖} = - 2.87 V$ ;

Reduction :

$H_{2} O_{2} + 2 H^{\oplus} + 2 e^{-} \to 2 H_{2} O$

$E^{⊖} = - 1.77 V$ ;

$2 I^{⊖} \to I_{2} + 2 e^{-}$ $E^{⊖} = - 0.54 V$ ;

Which of the following statements is/are correct ?

H_{2} O_{2} \to O_{2} + 2 H^{\oplus} + 2 e^{-}

E^{⊖} = - 0.69 V

H_{2} O_{2} + 2 H^{\oplus} + 2 e^{-} \to 2 H_{2} O

E^{⊖} = 1.77 V

I^{⊖} \to + 2 e^{-}

E^{⊖} = - 0.535 V

lim x \to 0 \frac{e (1 + x)^{2} + (e + x) ℓ n (e + x) - 2 e}{(1 - cos x + sin x)}

A^{2 +} + {2 e}^{-} ⟶ A; E^{o} = 0.34 V

X ⟶ X^{2 +} + {2 e}^{-}; E^{o} = 2.37 V

E_{c e l l}

Join BYJU'S Learning Program