coulmb's law

Can anybody help me to sort out " what is force between two charges Q1 and Q2 separeted by a distance of R between them, if up to a distance of R/2 , permittivity ε1 and in the other R/2 distance has permittivity ε2 medium.

higher permittivity ε2 can see a same as shorten distance of ε1 in electrical eye.

in situation ε1=1, ε2=2:

you need double distance between Q in ε2 medium for same force as ε1 medium


or you case can see electrostatic 'resistance' ε2 half of ε1 for same distance and high ε material more or less 'shorten' electrical field per distance and in bad situation with serial combine high ε and low ε material in elecrostatic field, can give very high field strengt over thin low ε material.

so convert R1/2 ε1 and R/2 ε2 to pure equal distance of ε1 give 1/2 ε1 + 1/4 ε2 => 3/4 ε1

ie. you calculate force by R3/4 of ε1 in your formula.

(possibly can thinking wrong here...)

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for example cavity filtes with very high power, can make corona in air on resonator surface depend of very high fieldstrength on small area, and one method to protect this is using thin plastic layer (high ε) on resonator surface to transporting out electrical field strength from metallic surface to bigger area and softer edge to air, ie transforming metallic surface permittivity to air permittivity with more smaller step, not one big step.

same reason used for very high voltage cable and structures...