Question:
An electromagnetic wave travels in a medium with a speed of \( 2 \times 10^8 \, \text{ms}^{-1} \). The relative permeability of the medium is 1. Then the relative permittivity is.
An electromagnetic wave travels in a medium with a speed of \( 2 \times 10^8 \, \text{ms}^{-1} \). The relative permeability of the medium is 1. Then the relative permittivity is.
Show Hint
To find the relative permittivity of a medium, use the relationship between the speed of light in the medium and the speed of light in vacuum.
Updated On: Mar 16, 2026
- 1.75
- 2
- 2.25
- 2.75
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
The speed of light in a medium is given by:
\[
v = \frac{c}{\sqrt{\mu_r \epsilon_r}}
\]
Where:
- \( v \) is the speed of the electromagnetic wave in the medium
- \( c \) is the speed of light in a vacuum
- \( \mu_r \) is the relative permeability of the medium
- \( \epsilon_r \) is the relative permittivity of the medium
Given that \( \mu_r = 1 \) and \( v = 2 \times 10^8 \, \text{ms}^{-1} \), and \( c = 3 \times 10^8 \, \text{ms}^{-1} \), we can solve for \( \epsilon_r \):
\[
\epsilon_r = \frac{c^2}{v^2} = \frac{(3 \times 10^8)^2}{(2 \times 10^8)^2} = 2.25
\]
Thus, the relative permittivity is \( 2.25 \).
Was this answer helpful?
1
0
Top AP EAMCET Physics Questions
- If only $\frac{1}{51}^{th}$ ot the main current is to be passed through a galvanometer then the shunt required is $R_1$ and if only $\frac{1}{11}^{th}$ of the main voltage is to be developed across the galvanometer, then the resistance required is $R_2$. Then $\frac{R_2}{R_1} =$
In the given circuit, the electric currents through $15\, \Omega$ and $6 \, \Omega$ respectively are
- In a nuclear reactor the activity of a radioactive substance is $2000 / s$. If the mean life of the products is $50$ minutes, then in the steady power generation, the number of radio nuclides is
- In a $p$ - type semiconductor the donor level is at $50 \,meV$ above the valence band. To produce one electron, the maximum wavelength of light photon required is (Planck's constant, $h=6.6 \times 10^{-34} \,Js$ and speed of light in vacuum, $c=3 \times 10^{8} \,ms ^{-1}$ )
- Potential energy of a body of mass $1\, kg$ free to move along X-axis is given by $U(x) = \left( \frac{x^2}{2} - x \right) J .$ If the total mechanical energy of the body is $2\, J$. then the maximum speed of the body is (Assume only conservative force acts on the body)
View More Questions
Top AP EAMCET Electromagnetic induction Questions
- In a circuit, the current falls from 14 A to 4 A in a time 0.2 ms. If the induced emf is 150 V, then the self-inductance of the circuit is:
- An electron is moving with a velocity \( (2\hat{i} + 3\hat{j}) \) m/s in an electric field \( (3\hat{i} + 6\hat{j} + 2\hat{k}) \) V/m and a magnetic field \( (2\hat{j} + 3\hat{k}) \) T. The magnitude and direction (with x-axis) of the Lorentz force acting on the electron is
A magnet suspended in a uniform magnetic field is heated so as to reduce its magnetic moment by 19%. By doing this, the time period of the magnet approximately
- If the current through an inductor increases from 2 A to 3 A, the magnetic energy stored in the inductor increases by
- The Solar Radiation is
View More Questions
Top AP EAMCET Questions
- If $\int \cos x . \cos 2x . \cos 5x dx = A \; \sin 2x + B \sin 4x + C \sin 6x + D \sin 8x + k $ (where $k$ is the arbitrary constant of integration), then $\frac{1}{B} + \frac{1}{C} = $
- If $k$ is one of the roots of the equation $x^2 - 25x + 24 = 0 $ such that $A = \begin{bmatrix}1&2&1\\ 3&2&3\\ 1&1&k\end{bmatrix} $ is a non-singular matrix, then $A^{-1}$ =
- If only $\frac{1}{51}^{th}$ ot the main current is to be passed through a galvanometer then the shunt required is $R_1$ and if only $\frac{1}{11}^{th}$ of the main voltage is to be developed across the galvanometer, then the resistance required is $R_2$. Then $\frac{R_2}{R_1} =$
- If $p$ and $q$ are respectively the global maximum and global minimum of the function $f(x) = x^2 e^{2x}$ on the interval $[-2, 2]$ , then $pe^{-4} + qe^4 = $
- If $\displaystyle\sum^n_{k - 1} \tan^{-1} \left( \frac{1}{k^2 + k + 1} \right) = \tan^{-1} (\theta) $ , then $\theta$ =
View More Questions