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The ideal parallel capacitor has been filled with 2 dielectrics as shown in the above diagram.

$$ V_{1} :=\text{potential at the top plate} $$

$$ V_{2} :=\text{potential at the bottom plate} $$

$$ V' :=\text{ potential at the border between the dielectrics } $$

$$ V :=\text{ potential at any point inside the any colored domain } $$

We want to deduce the formula of \$V\$

$$ D :=\text{ electric flux density } ~~ \leftarrow~~ \text{this value is constant at any point between the plates.} $$

$$ E_{1}=\frac{ D }{ \epsilon_{1} } $$

$$ E_{2}=\frac{ D }{ \epsilon_{2} } $$

The below 2 equations are the problems for me currently.

$$ V'-V_{1} = E_{1} d_{1}$$

$$ V_{2} -V'= E_{2} d_{2} $$

Concretely I can't get relations of magnitudes of the each potential.

Since

$$ E_{1} d_{1} ~~,~~ E_{2} d_{2} >0 $$

The below inequations must be held.

$$ V_{2} \geq V' \geq V_{1} \geq 0 $$

How can it be determined?

Why \$V_{2}\$ can be asserted that \$V_{2}\$ takes a highest value?

I thought the below.

$$ Q:=\text{charge at the top plate} $$

$$ \sigma :=\text{surface charge density of the top plate} $$

Since the electric flux density \$D\$ is constant at any point between the plates,

the surface charge density of the bottom plate can be said \$ -\sigma \$

By the way and needless to say , \$V'\$ is constant at any point inside the surface of border of the 2 dielectrics.

Of course electric lines of force mostly exist between the plates(few of electric lines of force exist outside between the plates).

What I've been missing or mistaking?

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When you assumed $$ E_{1} d_{1} ~~,~~ E_{2} d_{2} >0 $$ you assumed the electric field points upwards. Therefore you assumed the bottom plate is at a higher potential than the top plate.

However, I don't see any reason you should assume the field points upwards.

It's probably better if you assume \$V_1\$ and \$V_2\$ are known values and use them to calculate \$E_1\$ and \$E_2\$. You might even decide that either \$V_1\$ or \$V_2\$ is taken to be 0 (i.e. is the reference potential) to simplify the calculations.

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