Question

A parallel plate capacitor of capacitance Co has plates of area A with separation d between them. When it is connected to a battery of voltage Vo it has charge of magnitude Oo on its plat It is then disconnected from the battery and the space between the plates is filled with a material of dielectric constant 3 After is added, the magnitudes of the charge on the plates the dielectric and the potential change how? You of course must justify your conclusions

I need correct answers. If you do not know how to solve it please do not answer, or copy other similar answers. I will rate the correct answer.

Answer 1

Capacitance C of a parallel palte capacitor is given by C = KeoA/d

where A = area = pi r^2,

e0 = constnat = 8.85*10^-12,

d   = distance between the plates,

K = dieelctric constant (=1 for air)

Chareg Q = CV where V = Volatge

so here

if K is introduced

Q = KCV

so if K is introduced Charge increases

and V = Vo/K

voltgae decreases