To determine the intensity of the plane electromagnetic wave described by \( E_y = (200 \, \text{V/m}) \sin(1.5 \times 10^7 t - 0.05 x) \), we need to use the formula for the intensity of an electromagnetic wave:
The intensity \( I \) of an electromagnetic wave is given by the formula:
\(I = \frac{1}{2} \varepsilon_0 c E_0^2\)
Where:
Substituting these values into the formula, we get:
\(I = \frac{1}{2} \times 8.85 \times 10^{-12} \, \text{F/m} \times 3 \times 10^8 \, \text{m/s} \times (200 \, \text{V/m})^2\)
Calculating step-by-step:
\(I = \frac{1}{2} \times 2.655 \times 10^{-3} \times 40000\)
\(I = 1.3275 \times 10^{-3} \times 40000\)
\(I = 53.1 \, \text{W/m}^2\)
This matches the option 53.1 W/m², confirming it as the correct answer.
The intensity \(I\) of an electromagnetic wave is given by:
\[ I = \frac{1}{2} \epsilon_0 c E_0^2 \]
where \(\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2\), \(c = 3 \times 10^8 \, \text{m/s}\), and \(E_0 = 200 \, \text{V/m}\).
Substituting the values:
\[ I = \frac{1}{2} \times 8.85 \times 10^{-12} \times (3 \times 10^8) \times (200)^2 \] \[ I = 53.1 \, \text{W/m}^2 \]
The dimension of $ \sqrt{\frac{\mu_0}{\epsilon_0}} $ is equal to that of: (Where $ \mu_0 $ is the vacuum permeability and $ \epsilon_0 $ is the vacuum permittivity)
A point particle of charge \( Q \) is located at \( P \) along the axis of an electric dipole 1 at a distance \( r \) as shown in the figure. The point \( P \) is also on the equatorial plane of a second electric dipole 2 at a distance \( r \). The dipoles are made of opposite charge \( q \) separated by a distance \( 2a \). For the charge particle at \( P \) not to experience any net force, which of the following correctly describes the situation?

