Question

In a parallel plate capacitor connected to a constant voltage source, a dielectric is introduce such that its surface area is same as that of plates but the thickness of the dielectric is half the separation between the plates. If dielectric constant of the dielectric between the plates is 6, then

• A. percentage loss of energy of the capacitor, for introducing the dielectric between the plates is 41.67%
• B. the capacitors decreases by about 71%
• C. the potential difference across the plates of the capacitors decreases by about 71%
• D. the charge of the capacitor remains conserved while introducing the dielectric between the plates.

Answer 1

Answer: (D)

the charge of the capacitor remains conserved while introducing the dielectric between the plates.

Initially the capacitance $C=\frac{A{ϵ}_0}{d}$
When dielectric fill the half separation $\left(d/2\right)$ of plates, there are two capacitors- one is air filled with capacitance $2C$ and another is dielectric filled with capacitance $2kC$ and they are in series.
$C_{eq}=\frac{2C.2kC}{2C+2kC}=\frac{2k}{k+1}C=\frac{2\left(6\right)}{6+1}C=\frac{12}{7}C=1.71C$
thus here capacitance will increase by $(1.71-1)100=71\%$
As the capacitor is connected to voltage source so total potential remain constant and energy stored in capacitors will also increase. $(U=1/2C_{eq}{V}^2)$
In a capacitor charge is always conserved and it will flow to maintain constant potential.