Brilliant Electromagnetism Contest

Since no one has posted a solution in 16 hours I'll post the solution.

Let $V_1 = 2 V \\ V_2 = 20 V \\ V_3 = 200 V \\ R_0 = 10 \Omega \\ i = 1 \, mA$

Then, we get that $i=\frac{V_k}{\sum_{j=0}^{k} R_j } \quad k = 1,2,3$

Solving, we get $R_1 = 1990 \Omega \\ R_2 = 18 k\Omega \\ R_3 = 180 k\Omega$

Thanks for maintaining rules and regulations of the contest. I'm inviting certain people to the contest, and if they don't respond, I shall declare the end of the same.

Let , mass of proton $= m$, mass of deuteron = $2m$, Magnetic field = $B$

Charge of both particles $= q$, velocity of proton = ${v_{p}}$, velocity of deuteron = ${v_{d}}$

As they are accelerated to same potentials and their kinetic energy are same(if we consider the given potential energy is completely transferred to kinetic energy, then they always have same kinetic energy because they both have same charge), so

$m{v_{p}}^2 = 2m{v_{d}}^2$ $\implies$ $v_{p} = \sqrt{2} v_{d}$

The magnetic field is same for both the particles. As they trace the circular path, their centripetal force will be equal to magnetic force which are in opposite direction because the their velocity and magnetic field are both perpendicular.

So, $qvB = \dfrac{m{v}^2}{r}$ $\implies$ $r=\dfrac{mv}{qB}$,

As , $q, B$ are same for both, $r∝ mv$

$∴ \dfrac{r_{p}}{r_{d}} = \dfrac{m{v_{p}}}{2m{v_{d}}}$

$∴ r_{d} = \sqrt{2} r_{p}$

is the answer $\sqrt{2}$ ${r}_{p}$ ? @Swapnil Das

$E=$ Electric Field at point P lying inside sphere

Let the sphere be made of infinite uniformly charged shells with radius varying from 0 to R and charges $q_1 , q_2 ,q_3 ,...,q_{R}$ and charge density $\rho$

$E = E_{inside shells} + E_{outside shells}$

= $E_{ inside shells}$

=$\dfrac{Kq_1}{r^{2}}+\dfrac{Kq_2}{r^{2}}+...+\dfrac{Kq_r}{r^{2}}$

=$\dfrac{Kq_{inside}}{r^{2}}$

=$\dfrac{1}{4π\epsilon_0r^{2}} \times \rho \times \dfrac{4πr^{3}}{3}$

=$\dfrac{\rho r }{3\epsilon_0}$

Let the centre of sphere & cavity be $O$ and $P$ respectively. The sphere with cavity is equivalent to system of 2 spheres , one sphere of Radius R & charge density $\rho$ and another sphere of radius $r$ & charge density $-\rho$ coinciding with the cavity.

$E_1 = E_{\rho}+E_{-\rho}$

=$\dfrac{\rho r}{3\epsilon_0}+0$

$\dfrac{E}{E_1}=\boxed{1}$

Is the answer 1?

Is the answer ((R/r)^3-1)^-1?

$\begin{aligned} dV &= \frac{1}{4 \pi \epsilon_0} \frac{\lambda_0 \cos \frac{\theta}{2} R\, d\theta}{R} \\ \implies V &= \frac{1}{4 \pi \epsilon_0} \int_{0}^{2 \pi } \lambda_0 \cos \frac{\theta}{2} \, d\theta \\ &= \boxed{0} \end{aligned}$

As total charge on the ring is,

$dq = \lambda_{0} cos(\dfrac{\theta}{2})$

$Q = \int_{0}^{2\pi} \lambda_{0} cos(\dfrac{\theta}{2}) d\theta = 0.$

So potential is also zero at anywhere.