Problem set 11

1. An electron with rest mass $m_0$ is accelerated. Its relativistic mass is $2m_0$ when its speed is
1. $c$
2. $\frac{c\sqrt{3}}{2}$
3. $c\sqrt{3}$
4. $2c$
Relativistic mass is given by $m=\frac{m_0}{\sqrt{1-v^2/c^2}}$, substituting $m=2m_0$ and solving we get $v=\frac{c\sqrt{3}}{2}$

2. Twelve equal charges of magnitude $q$ are kept at the corners of a regular 12-sided polygon one at each of the corner. What is the net force on a test charge $Q$ at the center of the polygon at distance $r$?
1. Zero
2. $\frac{1}{4\pi\epsilon_0}\frac{12qQ}{r^2}$
3. $\frac{1}{4\pi\epsilon_0}\frac{6qQ}{r^2}$
4. $\frac{1}{4\pi\epsilon_0}\frac{qQ}{r^2}$
As charge $Q$ is placed at symmetric position the net force acting is zero.

3. Two equal charges each of magnitude $q$ are kept $d$ distance apart along X-axis. Now, the electric field at a distance $z(z>>d)$ above the midpoint between two charges is given by:

1. $E_z=\frac{1}{4\pi\epsilon_0}\frac{2q}{z^2}$
2. $E_z=\frac{1}{4\pi\epsilon_0}\frac{q^2}{z^2}$
3. $E_z=\frac{1}{4\pi\epsilon_0}\frac{2q}{r^{3/2}}$
4. $E_z=\frac{1}{4\pi\epsilon_0}\frac{2q}{(2+r)^2}$
Electric field at a point at a distance $z$ due to single charge is $E_1=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$. Hence electric field at the same point along z-axis is $E_1=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\cos\theta$. But $\cos\theta=\frac{z}{r}$ and $r=z^2+\frac{d^2}{4}$. Hence \begin{align*} E_1&=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\frac{z}{r}\\ &=\frac{1}{4\pi\epsilon_0}\frac{qz}{r^3}\\ &=\frac{1}{4\pi\epsilon_0}\frac{qz}{(z^2+\frac{d^2}{4})^{3/2}} \end{align*} For $z>>d$, $\frac{d^2}{4}$ can be neglected. Hence,$E_1=\frac{1}{4\pi\epsilon_0}\frac{q}{z^2}$. Similarly, $E_2=\frac{1}{4\pi\epsilon_0}\frac{q}{z^2}$. Hence, $E=E_1+E_2=\frac{1}{4\pi\epsilon_0}\frac{2q}{z^2}$

4. Poynting's vector represent which of the following quantities?
   1. Current density vector producing electrostatic field
   2. Power density vector producing electromagnetic field
   3. Current density vector producing electromagnetic field
   4. Power density vector producing electrostatic and magnetostatic fields
   (B) Power density vector producing electromagnetic field.
   
5. An n-type semiconductor has an electron concentration of $3\times 10^{20} m^{-3}$. If the electron drift velocity is $100 ms^{-1}$ in an electric field of $200 Vm^{-1}$, the conductivity (in $\Omega m^{-1}$)
   1. 24
   2. 36
   3. 48
   4. 96
   Mobility $$\mu=\frac{v_d}{E}$$ Conductivity \begin{align*}\sigma&=n_e e\mu\\&=\frac{n_e e v_d}{E}\\&=\frac{3\times 10^{20}\times1.6\times 10^{-19}\times100}{200}\\&=24\end{align*}