Two parallel plate capacitors differ only in the spacing between their (very thin) plates , AB has a spacing of `5` mm and a capacitance of `20 pF`, while CD has a spacing of `2 mm` . Plates `A` and `C` carry charges of `+1 nC`, while B and D each carry `-1 nC`, What are the potential diffences ` V_(AB)` and `V_(CD)` after the capacitor `CD` is slid centrally between and parallel ot the plates of AB without touching them ? Would it make any difference if CD was not centrally paced between A and B ?
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Explanation

The net charges on the plates cannot change, but the charges on the plates on either of any of spaces must be equal and opposite. Consequently, the charges on `C` and `D` must be `-1nC` and `+nC`, respectively, on their outside surfaces and` +2nC` and `-2 nC`, respectively, on their surfaces. The capacitance of any pair plates is inversely proportional to their separation,with `5 mm` corresponding to `20 pF`. Thus if `AC` is `x mm` and `DB` is `(3-x) mm`, the capacitances of the three successive capacitors are `100//x, 50`, and `100//(3-x)pF`. The valtage `V_(CD)` is therefore, `40 V`, and `V_(AB)` is `10x+40+10(3-x)=70 V`, independent of the value of `x`.