Explanation:

Capacitance of a Parallel Plate Capacitor:
The capacitance of a parallel plate capacitor is given by the formula:

C = ε₀A/d

where C is the capacitance, ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.

Effect of Aluminum Foil:
When a highly conducting sheet of aluminum foil is placed between the plates of a parallel plate capacitor, it acts as an additional conducting plate. This modifies the electric field and capacitance of the capacitor.

Parallel Plate Capacitor with Aluminum Foil:
When the aluminum foil is inserted parallel to the plates of the capacitor, it divides the capacitor into two separate capacitors. One capacitor is formed between one plate and the foil, and the other capacitor is formed between the foil and the second plate.

Equivalent Capacitance:
The capacitance of capacitors in parallel is given by the formula:

C_eq = C₁ + C₂

where C_eq is the equivalent capacitance and C₁ and C₂ are the capacitances of the individual capacitors.

Effect on Capacitance:
Before inserting the aluminum foil, the capacitance of the parallel plate capacitor is given as 10 F.

After inserting the foil, the equivalent capacitance is the sum of the capacitance between one plate and the foil (C₁) and the capacitance between the foil and the other plate (C₂).

Since the aluminum foil is highly conducting, it acts as a perfect conductor. This means that the electric field inside the foil is zero. As a result, the electric field between the foil and each plate is the same as the electric field between the original plates.

Therefore, the capacitance between each plate and the foil is the same as the original capacitance of the parallel plate capacitor, which is 10 F.

Hence, the equivalent capacitance after inserting the foil is:

C_eq = C₁ + C₂ = 10 F + 10 F = 20 F

Conclusion:
The capacitance of the parallel plate capacitor increases to 20 F after inserting the highly conducting aluminum foil between the plates.