- #36

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And here is how you could have done it if you considered the "three capacitors in ~~parallel~~ series" approach that I suggested in post #2. One can use this approach because the equipotentials in this problem are planes parallel to the plates which is exactly the case with capacitors in series. Moving the dielectric to touch one of the plates is equivalent to swapping the order in the series capacitors.

Let the vacuum capacitances be ##C_1=C_3=C## and the dielectric capacitance be ##C_2=\epsilon_rC=2C.## The equivalent capacitance is $$\frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{2C}+\frac{1}{C}=\frac{5}{2C}\implies C_{eq}=\frac{2}{5}C.$$The common charge is ##Q=C_{eq}U_0=\dfrac{2}{5}CU_0.##

Then $$\begin{align} & V_1=V_3=\frac{Q}{C}=\frac{2}{5}U_0\implies E_{vac}=\frac{2}{5}\frac{U_0}a \nonumber \\ & V_2=\frac{Q}{2C}=\frac{1}{5}U_0\implies E_{diel}=\frac{1}{5}\frac{U_0}a. \nonumber \end{align}$$

Let the vacuum capacitances be ##C_1=C_3=C## and the dielectric capacitance be ##C_2=\epsilon_rC=2C.## The equivalent capacitance is $$\frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{2C}+\frac{1}{C}=\frac{5}{2C}\implies C_{eq}=\frac{2}{5}C.$$The common charge is ##Q=C_{eq}U_0=\dfrac{2}{5}CU_0.##

Then $$\begin{align} & V_1=V_3=\frac{Q}{C}=\frac{2}{5}U_0\implies E_{vac}=\frac{2}{5}\frac{U_0}a \nonumber \\ & V_2=\frac{Q}{2C}=\frac{1}{5}U_0\implies E_{diel}=\frac{1}{5}\frac{U_0}a. \nonumber \end{align}$$

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