Answer :
The battery adds 10µC of charge to the capacitors.
We have a 10V battery connected in series to two capacitors of 6µF each. When one of the capacitors gets squished, halving the distance between its plates while still connected to the battery, the capacitance of this capacitor changes.
Capacitance, C, is given by the formula:
C = ε₀(A/d), where ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.
When the distance d is halved, the capacitance C doubles. The new capacitance for the squished capacitor will be 12µF.
In a series circuit, the charge Q on each capacitor is the same. The total voltage V across the series combination of capacitors is given by:
- 1/C(total) = 1/C₁ + 1/C₂
Initially,
- 1/C(total) = 1/6 + 1/6 = 1/3 µF,
- C(total) = 3µF.
The total charge Q is given by:
- Q = C(total)* V = 3µF * 10V = 30µC
After squishing one capacitor to 12µF:
- 1/C(total)= 1/12 + 1/6 = 1/4 µF,
- C(total)= 4µF.
The new total charge Q is:
- Q = C(total)* V
- = 4µF * 10V
- = 40µC.
