A parallel plate capacitor of area A and separation d is fully charged to potential difference V and then removed from charging source. A dielectric slab of dielectric constant K=2, thickness d and area A/2 is inserted as shown in figure. Let σ1 be the free charge density at the conductor-dielectric surface, σ2 be the charge density at the conductor- vacuum surface. The value of new capacitance and potential difference respectively are

A parallel plate capacitor of area A and separation d is fully charged

1.	A parallel plate capacitor of area A and separation d is fully charged to potential difference V and then removed from charging source. A dielectric slab of dielectric constant K=2, thickness d and area A/2 is inserted as shown in figure. Let σ1 be the free charge density at the conductor-dielectric surface, σ2 be the charge density at the conductor- vacuum surface. The value of new capacitance and potential difference respectively are
Answer» A parallel plate capacitor of area $A$ and separation $d$ is fully charged to potential difference $V$ and then removed from charging source. A dielectric slab of dielectric constant $K = 2$ , thickness $d$ and area $A / 2$ is inserted as shown in figure. Let $σ_{1}$ be the free charge density at the conductor-dielectric surface, $σ_{2}$ be the charge density at the conductor- vacuum surface. The value of new capacitance and potential difference respectively are

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