Electromagnetic Field Puzzle Time!(SOLUTION)

The solution is simple if you think of the method of image charges here

First we have an image point charge q2 so we can make the potential of the sphere that comes from q1 and q2 equal to 0

But we have a charge qs on the sphere so in order the sphere to have the same potential we must place a point charge q3 on its center with

q2+q3=qs

the place of q2 is b=r0^2/d = 1 cm

and its charge is q2= - ro/d * q1= 1nC

so q3= 0.5 nC

so we have 3 point charges in P1(0,0,0) P2(3,0,0) P3(4,0,0)

the distances of each point charge from the point P(1,1,1) is :

r1= sqrt 3 cm

r2= sqrt 6 cm

r3= sqrt 11 cm

so the potential is calculated :

Φ= 1/4πεrε0 * (q1/r1 + q2/r2 + q3/r3) = -214,16 V ( well its -214,158 but i said round it a a little )

That's the solution

Only 1 solver though and not with enough contribution value :/

Original problem :

Welcome to an electromagnetic puzzle!I hope its the first here in steamgifts :)

The giveaway is Fable:The Lost Chapters game (contributor of $40 required).

The problem is :

In a x,y,z 3D space we have the point charge q1 placed in (0(cm),0(cm),0(cm)) with value q1= -2(nC). We also have a solid charged conducting sphere of radius r= 2 (cm) with qs= 1.5 (nC) and its center is in (4(cm),0(cm),0(cm)). The distance between the center of the sphere and q1 is d= 4 (cm). (nC stands for nano Coulomb and cm for centimeters)

You have to find the electric potential Φ in point P(1(cm),1(cm),1(cm)).

The permittivity οf the space is ε=2.5 and permittivity of air is the known constant ε0=8,854 * 10^-12 (F/m)

Here is a graph to help you with the puzzle :

graph

once you find the answer you place it here(plain number,round it up a little) and you get the link of the giveaway

Giveaway lasts for about a week from now!

So grab your pens and start solving some electromagnetic field math!

Have fun and dont hesitate to ask me anything :)

I hope the puzzle was understandable :)

EDIT: Of course leaking or sharing the giveaway is prohibited !