Two identical air-filled parallel-plate capacitors C1 and C2 are connected in series to a battery that has voltage V. The charge on each capacitor is Q0. While the two capacitors remain connected to the battery, a dielectric with dielectric constant K>1 is inserted between the plates of capacitor C1, completely filling the space between them.

Complete question:

Two identical air-filled parallel-plate capacitors C₁ and C₂ are connected in series to a battery that has voltage V. The charge on each capacitor is Qo. While the two capacitors remain connected to the battery, a dielectric with dielectric constant K > 1 is inserted between the plates of capacitor C1, completely filling the space between them. What is the charge on capacitor C1 after the dielectric is inserted? Express your answer in terms of K and Qo.

The charge on capacitor after the dielectric is inserted is Qo( k + 1)

Explanation:

Given;

two identical capacitor, C₁ and C₂

charge on each capacitor, Q = Qo

Voltage of the battery, = V

dielectric constant K > 1

After inserting dielectric constant K, the equivalent capacitance in parallel becomes;

C.equ. = KC₁ + C₂

Charge, on capacitor Qo = C.equ.V

Q = V(KC₁ + C₂)

Q = KC₁V + C₂V

since, the charge on each capacitor is Qo, then

Q = KQo + Qo

Q = Qo( k + 1)

Therefore, the charge on capacitor after the dielectric is inserted is Qo( k + 1), in terms of K and Qo.

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this