Aparallel-plate capacitor initially has a potential difference of 400 v and is then disconnected from the charging battery. if the plate spacing is now doubled (without changing q), what is the new value of the voltage?

The capacitance of a capacitor is the ratio of the stored charge to its potential difference, i.e.

C = Q/ΔV

C is the capacitance

Q is the stored charge

ΔV is the potential difference

Rearrange the equation:

ΔV = Q/C

We also know the capacitance of a parallel-plate capacitor is given by:

C = κε₀A/d

C is the capacitance

κ is the capacitor's dielectric constant

ε₀ is the electric constant

A is the area of the plates

d is the plate separation

If we substitute C:

ΔV = Qd/(κε₀A)

We assume the stored charge and the area of the plates don't change. Then if we double the plate spacing, i.e. we double the value of d, then the potential difference ΔV is also doubled.

increases it

i don't understand. explain it clearly