Problem set 20

1. Let $v$, $p$ and $E$ denote the speed, the magnitude of the momentum, and the energy of a free particle of rest mass $m$. Then
1. $dE/dp=constant$
2. $p=mv$
3. $v=cp/\sqrt{p^2+m^2c^2}$
4. $E=mc^2$
As particle is moving with velocity $v$ and having rest mass $m$, its relativistic mass is given by $$m_{rel}=\frac{m}{\sqrt{1-v^2/c^2}}$$ Its momentum is given by $$p=m_{rel}v=\frac{mv}{\sqrt{1-v^2/c^2}}$$ Squaring and rearranging this equation, we get, $$p^2\left(1-v^2/c^2\right)=m^2v^2$$ On solving this equation, we get, $v=cp/\sqrt{p^2+m^2c^2}$. Hence, option $(C)$ is correct.
The rest mass energy and kinetic energy of particle are $mc^2$ and $K=\frac{mc^2}{\sqrt{1-v^2/c^2}}-mc^2$, respectively.
Here, option (D) is wrong, because, total energy $E$, is sum of kinetic energy plus rest mass energy i.e. $E=\frac{mc^2}{\sqrt{1-v^2/c^2}}$.
Option (A) is wrong, because, velocity is involved in the expression for $E$.
2. A binary star system consists of two stars $S_1$ and $S_2$, with masses $m$ and $2m$, respectively, separated by a distance $r$. If both $S_1$ and $S_2$ individually follow circular orbits around the centre of mass with instantaneous speeds $V_1$ and $V_2$ respectively, the speeds ratio $V_1/V_2$ is
1. $\sqrt{2}$
2. 1
3. 1/2
4. 2
Let the stars are at distances $r_1$ and $r_2$ from centre of mass, hence, we have $$m_1r_1=m_2r_2\quad\text{and}\quad r_1+r_2=r$$ For the motion of first star in circular orbit, we have $$\frac{Gm_1m_2}{r}=\frac{m_1v_1^2}{r_1}$$ For the motion of second star in circular orbit, we have $$\frac{Gm_1m_2}{r}=\frac{m_2v_2^2}{r_2}$$ Taking ratio of these two equations, we get, $$\frac{v_1}{v_2}=\sqrt{\frac{r_1m_2}{r_2m_1}}$$ Substituting $\frac{r_1}{r_2}=\frac{m_2}{m_1}$ $$\frac{v_1}{v_2}=\frac{m_2}{m_1}=2$$ Hence, answer is (D)
3. Three charges are located on the circumference of a circle of radius $R$ as shown in the figure below. The two charges $Q$ subtend an angle $90^o$ at the centre of the circle. The charge $q$ is symmetrically placed with respect to the charges $Q$. If the electric field at the centre of the circle is zero, what is the magnitude of $Q$?

1. $q/\sqrt{2}$
2. $\sqrt{2}q$
3. $2q$
4. $2q$

From figure, for balancing, we must have $$E_q=2E_Q\cos{(45)}$$ i.e. $\frac{1}{4\pi\epsilon_0}\frac{q}{R^2}=2\frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}\frac{1}{\sqrt{2}}$. Hence, $Q=q/\sqrt{2}$.
4. Consider a hollow charged shell of inner radius $a$ and outer radius $b$. The volume charge density is $\rho(r) =\frac{k}{r^2}$ ($k$ is a constant) in the region $a < r < b$. The magnitude of the electric field produced at distance $r > a$ is
1. $\frac{k(b-a)}{\epsilon_0r^2}$ for all $r > a$
2. $\frac{k(b-a)}{\epsilon_0r^2}$ for $a < r < b$ and $\frac{kb}{\epsilon_0r^2}$ for $r > b$
3. $\frac{k(r-a)}{\epsilon_0r^2}$ for $a < r < b$ and $\frac{k(b-a)}{\epsilon_0r^2}$ for $r > b$
4. $\frac{k(r-a)}{\epsilon_0a^2}$ for $a < r < b$ and $\frac{k(b-a)}{\epsilon_0a^2}$ for $r > b$

Let us divide the region $r > a$ into two regions viz. $a\leq r\leq b$ and $r > b$.
For region $a\leq r\leq b$, using Gauss law, we have, $$\int\vec E\cdot d\vec s=\frac{q_{enclosed}}{\epsilon_0}$$ \begin{align*} E\;4\pi r^2&=\frac{1}{\epsilon_0}\int_a^r\rho\:dV\\ &=\frac{1}{\epsilon_0}\int_a^r\frac{k}{r^2}\:4\pi r^2\;dr\\ &=\frac{4\pi k}{\epsilon_0}(r-a) \end{align*} $$E=\frac{k(r-a)}{\epsilon_0 r^2}$$ For region $r > b$, using Gauss law, we have, $$\int\vec E\cdot d\vec s=\frac{q_{enclosed}}{\epsilon_0}$$ \begin{align*} E\;4\pi r^2&=\frac{1}{\epsilon_0}\int_a^b\rho\:dV\\ &=\frac{1}{\epsilon_0}\int_a^b\frac{k}{r^2}\:4\pi r^2\;dr\\ &=\frac{4\pi k}{\epsilon_0}(b-a) \end{align*} $$E=\frac{k(b-a)}{\epsilon_0 r^2}$$ Hence, answer is (C)
5. Consider the interference of two coherent electromagnetic waves whose electric field vectors are given by $\vec E_1 = \hat i E_0 \cos{(\omega t+\phi)}$ and $\vec E_2 = \hat j E_0 \cos{(\omega t+\phi)}$ where $\phi$ is the phase difference. The intensity of the resulting wave is given by $\frac{\epsilon_0}{2}\left < E^2\right >$, where $\left < E^2\right > $ is the time average of $E^2$. The total intensity is
1. $0$
2. $\epsilon_0E_0^2$
3. $\epsilon_0E_0^2\sin^2{\phi}$
4. $\epsilon_0E_0^2\cos^2{\phi}$
Resultant electric field is given by \begin{align*} \vec E &= \hat i E_0 \cos{(\omega t+\phi)}\\ &+\hat j E_0 \cos{(\omega t+\phi)} \end{align*} $$ E^2 = 2 E_0^2 \cos^2{(\omega t+\phi)}$$ \begin{align*} \left < E^2\right > &=\frac{\int_0^TE^2\,dt}{\int_0^T\,dt}\\ &=\frac{2E_0^2}{T}\int_0^T\cos^2{(\omega t+\phi)}\;dt\\ &={\scriptstyle\frac{2E_0^2}{T}\frac{1}{2}\int_0^T\left[1-\cos2{(\omega t+\phi)}\right]\;dt}\\ &=\frac{E_0^2}{T}T=E_0^2 \end{align*} $$I=\frac{\epsilon_0}{2}\left < E^2\right > =\frac{\epsilon_0E_0^2}{2}$$