Solving for Image Charges in Dielectrics

[Andy123]

Homework Statement

The question is attached.

Homework Equations

The Attempt at a Solution

I understand that the question would return to a typical conductor problem when ε2 >> ε1 so for the quadrant one, I placed image charges of q_1 at (d1 , 0 , -d2), -q1 at (-d1 , 0 , -d2) and another q1 at (-d1 , 0 , d2) with q1 = (ε1 - ε2)*q/(ε1 + ε2). When I applies the boundary conditions, I find that the image charges for quadrant two is q - q_1 at (d1 , 0 , d2), -q1 at (-d1 , 0 , -d2) and q1 at (-d1 , 0 , d2), for quadrant four is q - q_1 at (d1 , 0 , d2), -q1 at (-d1 , 0 , -d2) and another q1 at (-d1 , 0 , d2). The boundary conditions are well satisfies with these image charges. However, I failed to find image charges for quadrant three satisfying both the continuity of electric field and voltage at the boundaries connecting it with quadrant two and four. I have been working on this question for days and I am really frustrated. Is my attempt of approach is incorrect? Thank you for any help!

 

[TSny]

for the quadrant one, I placed image charges of q_1 at (d1 , 0 , -d2), -q1 at (-d1 , 0 , -d2) and another q1 at (-d1 , 0 , d2) with q1 = (ε1 - ε2)*q/(ε1 + ε2).

OK

When I applies the boundary conditions, I find that the image charges for quadrant two is q - q_1 at (d1 , 0 , d2), -q1 at (-d1 , 0 , -d2) and q1 at (-d1 , 0 , d2)

Are the q₁ and -q₁ necessary here? Can you satisfy the boundary conditions with just the q - q₁ charge?

When finding the potential or field inside the blue dielectric, I don't think you need to work with each of the three blue quadrants separately. See if you can treat the entire blue dielectric with just one image charge.