Question

A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?

Options

• The energy stored in the capacitor decreases K times.

• The change in energy stored is `1/2 "CV"^2 (1/"K" - 1)`.

• The charge on the capacitor is not conserved.

• The potential difference between the plates decreases K times.

Answer 1

The charge on the capacitor is not conserved.

Explanation:

Capacitance of the capacitor, C = `"Q"/"V"`

After inserting the dielectric, new capacitance

C' = K.C

New potential difference

V' = `"V"/"K"`; u_i = `1/2 "CV"^2 = "Q"^2/(2"C")` ...(∵ Q = CV)

u_f = `"Q"^2/(2"f") = "Q"^2/(2"kc") = ("C"^2"V"^2)/(2"KC") = ("u"_"i"/"k")`

Δu = u_f − u_i = `1/2 "CV"^2{1/"k" - 1}`

As the capacitor is isolated, so change will remain conserved p.d. between two plates of the capacitor

L = `"Q"/"KC" = "V"/"K"`