# [Physics] Why is the potential difference across two connected capacitors equal to the potential difference across each individual capacitor

capacitanceelectric-circuitshomework-and-exercisesvoltage

Quote from textbook:

The potential difference across the combination is the same as that across each individual capacitor.

Where, "the combination" refers to this situation : An $8 μF$ capacitor charged to a potential difference of $200 V$ and a $4 μF$ capacitor charged to a potential difference of $800 V$ are connected in parallel by joining terminals of like polarity.

My issue: Because the capacitors are connected in parallel and follow the rule $Ceq = C1 + C2$, and $Q=CV$, Hence $V=Q/C$ …

Wouldn't

$Ceq = C1 + C2$

Become

$Veq =$Q1/C1$+$Q2/C2 i.e. the sum of the individual potential differences.

Why would the potential difference across the combination of capacitors be the same as each of the individual capacitors?

FYI – for some figures:

Before connection ($Q=CV$):

$C1=8μF$, $V1=200V$, $Q1=1600 μC$

$C2=4μF$, $V1=800V$, $Q2=3200 μC$

After connection: (charge redistributes according to each capacitance):

$Q1=3200 μC$

$Q2=1600 μC$

Potential difference across the combination (according to textbook, which doesnt make sense to me as per the above question I asked): $V=Q/C=3200/8=400V$

Whereas I thought the potential difference across the combination would be = potential difference across capacitor $1$ + potential difference across capacitor 2 = $400+400=800V$

When you connect two capacitors in parallel, two things happen: (1) the capacitances $C1$ and $C2$ and the charges $Q1$ and $Q2$ add, i.e., $$C_p=C1+C2$$ and $$Q_p=Q1+Q2$$ (2) the voltage across the parallel connected capacitors has to be equal (both metallic terminals are each at equal potential) and becomes $$V_p=\frac{Q_p}{C_p}$$ Thus, in this particular case, the total charge is $Q_p=4800 \mu C$, the total capacitance is $C_p=12 \mu F$ so that the voltage is $V_p=400V$