Multitalented Multi - Conductors!

Method of images kills this problem.

Just we want a configuration where net potential on the conductors is zero.Clearly a configuration of charges placed placed on the vertices of square ABCD where adjacent vertices charges have equal magnitude and opposite nature.The equipotential lines here pass through the perpendicular bisectors of the sides of squares.Now just do the basic computations and our answer comes.

a=4

b=c=d=2

e=7

abcde = 224

My first lvl5 EM problem i solved on brilliant :P

I used the method of image charges ( may be the standard approach ).

In the picture above the red dots represent positive charge and the blue dots represent the negative charge.

Now from the Coulomb's law we have(as shown in the above picture):

$\large \vec{F_1}=\frac{-q^2}{4\pi\epsilon_0(2a)^2}a_y$

$\large \vec{F_2}=\frac{-q^2}{4\pi\epsilon_0(2b)^2}a_x$

$\large \vec{F_2}=\frac{-q^2}{4\pi\epsilon_0((2a)^2+(2b)^2)^{\frac{3}{2}}}(a_x2b+a_y2a)$

So now we have $\vec{F}=\vec{F_1}+\vec{F_2}+\vec{F_3}$

$\Rightarrow\large \vec{F}=\frac{q^2}{16\pi\epsilon_0}[a_x(\frac{b}{(a^2+b^2)^\frac{3}{2}}-\frac{1}{b^2})+a_y(\frac{a}{(a^2+b^2)^\frac{3}{2}}-\frac{1}{a^2})]$

$\Rightarrow|\vec{F}|=\frac{q^2}{16\pi\epsilon_0}\sqrt{[(\frac{b}{(a^2+b^2)^\frac{3}{2}}-\frac{1}{b^2})]^2+[(\frac{a}{(a^2+b^2)^\frac{3}{2}}-\frac{1}{a^2})]^2}$

$\Rightarrow -ma=\frac{q^2}{16\pi\epsilon_0}\sqrt{[(\frac{b}{(a^2+b^2)^\frac{3}{2}}-\frac{1}{b^2})]^2+[(\frac{a}{(a^2+b^2)^\frac{3}{2}}-\frac{1}{a^2})]^2}$

where $a$ is the acceleration and is given by:

$\Rightarrow a=-\frac{q^2}{16\pi\epsilon_0\cdot m}\sqrt{[(\frac{b}{(a^2+b^2)^\frac{3}{2}}-\frac{1}{b^2})]^2+[(\frac{a}{(a^2+b^2)^\frac{3}{2}}-\frac{1}{a^2})]^2}$

Also observe that the positive charge is symmetrically distant from both the infinite conductors so its path will be a straight line joining the point at which both the conductors meet. Hence we can take a=b=x(as given in the image). Putting this in the above equation we get

$\Rightarrow a=-\frac{q^2}{16\pi\epsilon_0\cdot m}\frac{(2\sqrt2-1)}{x^2}$

Now let $k=\frac{q^2}{16\pi\epsilon_0\cdot m}{(2\sqrt2-1)}$

We know $a=v\cdot\frac{dv}{dx}$ .

$\Rightarrow\large v\cdot\frac{dv}{dx}=-\frac{k}{x^2}$

$\Rightarrow\large v\cdot dv=-\frac{k}{x^2} \cdot dx$

$\Rightarrow\large \int v\cdot dv = -\int \frac{k}{x^2} \cdot dx$

Now integrating and applying the initial condition that at $x=1, v=0$

$\Rightarrow\large \frac{v^2}{2} = k(\frac{1}{x} -1)$

$\Rightarrow\large v =- \sqrt{2 k(\frac{1}{x} -1)}$

$\Rightarrow\large \frac{dx}{dt} =- \sqrt{2 k(\frac{1}{x} -1)}$

$\Rightarrow\large t_f =\int^0_1\frac{-1}{\sqrt{2 k(\frac{1}{x} -1)}}\cdot dx$

$\Rightarrow\large t_f =\int^1_0\frac{1}{\sqrt{2 k(\frac{1}{x} -1)}}\cdot dx$

Computing the integral we have:

$\Rightarrow\large t_f =\frac{\pi}{2^{\frac{3}{2}}\cdot \sqrt k}$

$\Rightarrow \large t_f={\sqrt{ \frac{ 4 \sqrt{2} \pi^3 m \epsilon_° ( 2 \sqrt{ 2} + 1)}{ 7 q^2} }}$

Therefore $a=4,b=c=d=2,e=7$ .

$\Rightarrow a\times b\times c\times d \times e=\boxed{224}$

My 1st LEVEL 5 EM 400 points problem.