Answer
$V_1 = 10.0~V$
Work Step by Step
The equivalent capacitance of $C_1$ and $C_6$ is $3.00~\mu F+3.00~\mu F$ which is $6.00~\mu F$
As we found in part (b), the total charge on these two capacitors is $60.0~\mu C$
We can find the potential difference:
$V = \frac{q}{C}$
$V = \frac{60.0~\mu C}{6.00~\mu F}$
$V = 10.0~V$
The potential difference across both $C_1$ and $C_6$ is $10.0~V$
Therefore, $V_1 = 10.0~V$
