Two parallel plate capacitors `8.0 mu F` each , are connected in parallel to a 10V battery. One of the capacitors is then squeezed so that its plate separation is 50.0% of its initial value. Because of the squeezing, (a) how much additional charge is transferred to the capacitors by the battery and (b) what is the increase in the total charge stored on the capacitors?

Explanation

To solve the problem, we need to analyze the situation step by step.

### Step 1: Calculate the initial capacitance of the capacitors
Given that both capacitors are identical and have a capacitance of \( C_1 = C_2 = 8 \, \mu F \).

Since they are connected in parallel, the total initial capacitance \( C_{initial} \) is given by:
\[
C_{initial} = C_1 + C_2 = 8 \, \mu F + 8 \, \mu F = 16 \, \mu F
...