SM381 - September 2023

General remarks: the solutions contained on this page have been provided by former SMT359/SM381 students according to their best endeavours. Users are encouraged to make amendments, comment on subjective points, make suggestions, and develop solutions for questions not solved yet.
= Part 1 =
== Question 1 ==
The electrostatic force \textbf{F} between two charges, from Coulomb's law is
: \mathbf{F}_{12} = \dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{r^2_{12}}\mathbf{\widehat{r}}_{12}= \dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{|\mathbf{r}_1-\mathbf{r}_2|^3}(\mathbf{r}_1-\mathbf{r}_2),
and hence the total electric field \mathbf{E} at the origin O on a charge q_1 is given by
:\mathbf{E(r)}=\mathbf{F}/q_1=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_2}{|\mathbf{r}_1-\mathbf{r}_2|^3}(\mathbf{r}_1-\mathbf{r}_2).
So the total electric field at the origin due to the point charges q at A and -q at B is
:
\begin{align}
\mathbf{E} &= \frac{1}{4 \pi \epsilon_0} \frac{q}{(\sqrt{b^2+a^2})^3}\left(-b\ \mathbf{e}_x-a\ \mathbf{e}_y\right)
 +\frac{1}{4 \pi \epsilon_0} \frac{-q}{(\sqrt{b^2+a^2})^3}(-b\ \mathbf{e}_x +a\ \mathbf{e}_y)\\
 &= \frac{2aq}{4 \pi \epsilon_0(b^2+a^2)^{3/2}}\left(-\mathbf{e}_y\right)
\end{align}
== Question 2 ==
The total electromagnetic force acting on the point charge q can be found using the Lorentz force law
:
\begin{align}
\mathbf{F}&=q(\mathbf{E}+\mathbf{v\times B})\\
&=q(v\ \mathbf{e}_x\times B_0\ \mathbf{e}_z)\\
&=vqB_0\ (-\textbf{e}_y)
\end{align}
For the charge's velocity to remain \mathbf{v}=v\ \mathbf{e}_x, there must be no net force on the charge, so the force that must be applied is
:\mathbf{F}_1=vqB_0\ \textbf{e}_y
== Question 3 ==
The electrostatic potential difference \Delta V=V_A-V_B between point A at (R,\pi/2,0) and point B at (R,0,0) is 
:
\begin{align}
\Delta V &= -\int\mathbf{E}\cdot\mathrm d\mathbf{l}\\
&= -\int_{(r=R,\theta=\pi/2,\phi=0)}^{(r=R,\theta=0,\phi=0)}\left(a\dfrac{\sin\theta}{r^2}\mathbf{e}_r-a\dfrac{\cos\theta}{r^2}\mathbf{e}_\theta\right)\cdot\mathbf{e}_\theta\ \mathrm d\theta\\
&=\int_{\theta=\pi/2}^{\theta=0}a\dfrac{\cos\theta}{R^2}\ \mathrm d\theta\\
&= \left[a\dfrac{\sin\theta}{R^2}\right]_{\theta=\pi/2}^{\theta=0}=-\dfrac{a}{R^2}
\end{align}
== Question 4 ==
From the no-monopole law we know that all magnetic fields are divergence-free. So \mathbf{K} could be a magnetic field if div \mathbf{K} is equal to zero.
:
\begin{align}
 \text{div }\mathbf{K} &= \dfrac{1}{r}\dfrac{\partial}{\partial r}(rK_r)+\dfrac{1}{r}\dfrac{\partial K_\phi}{\partial\phi}+\dfrac{\partial K_z}{\partial z}\\
&=\dfrac{1}{r}\dfrac{\partial}{\partial r}(ra_rz)+\dfrac{\partial\left(a_z\dfrac{z^2}{r}\right)}{\partial z}\\
&=\dfrac{a_rz}{r}+\dfrac{2a_zz}{r}
\end{align}
Hence \mathbf{K} could be a magnetic field, if a_r=-2a_z