A medium having dielectric constant K > 1 fills the space between the plates of a parallel plate capacitor. The plates have large area, and the distance between them is d . The capacitor is connected to a battery of voltage V . as shown in Figure (a). Now, both the plates are moved by a distance of d 2 from their original positions, as shown in Figure (b). In the process of going from the configuration depicted in Fi

Step-by-step solution

Capacitance of both capacitors are C a = K ε 0 A d and C b = ε 0 A d + d K = ε 0 A K d K + 1 Now, charge stored is q a = K ε 0 A d V and q b = ε 0 A K V d K + 1 . Electric field between each parallel plate capacitor is E a = q a K A ε 0 = K ε 0 A V d K ε 0 A = V d and E b dielectric = E air K = q b K A ε 0 = ε 0 A K V d K + 1 K A ε 0 = V d K + 1 So, from above relations, we can see that capacitance decrease by a factor of 1 K + 1 . Work done in the process W =