Test: The Parallel Plate Capacitor (NCERT)

In a parallel plate capacitor the capacity of capacitor.

The capacity of capacitor increases if area of the plate is increased.

The magnitude of the electric field between the plates is F = σ/2ε₀ - (- σ/2ε₀) - σ/ε₀

Capacitance of a parallel plate capacitor
C = ε₀A/d ...(i)
Also capacitance = potential difference/charge .......(ii)
When battery is disconnected and the distance between the plates of the capacitor is increased then capacitance increases and charge remains constant.
Since capacitance = potential difference/charge
∴ Potential difference increases.

As the capacitor is isolated after charging, charge on it remains constant. Plate separation increases d, decreases C = ϵ₀A/d and hence increases potential V = q/C.

When key K is kept closed, condenser C is charged to potential V. When plates of capacitors are moved apart, its capacitance, C = ϵ_oA/d decreases.
As potential of condenser remains same, charge Q = CV decreases. So option is correct. Once key K is closed, condenser gets charged, Q = CV.
Now, if key K is opened, battery is disconnected, no more charging can occur i.e. Q remains same.
As plates or capacitor are moved apart, its capacity C = ϵ_oA/d decreases.
Therefore, its potential, V = q/C increases