A parallel plate capacitor with area 200 cm² and separation between the plates 1.5 cm, is connected across a battery of emf V. If the force of attraction between the plates is $$25 \times {10^{ - 6}}N,$$ the value of V is approximately : $$\left( {{ \in _o} = 8.85 \times {{10}^{ - 12}}{{{C^2}} \over {N.{m^2}}}} \right)$$

A capacitor C₁ = 1.0 $$\mu $$F is charged up to a voltage V = 60 V by connecting it to battery B through switch (1). Now C₁ is disconnected from battery and connected to a circuit consisting of two uncharged capacitors $${C_2} = 3.0\mu F$$ and C₃ = 6.0 $$\mu $$F through switch (2), as shown in the figure. The sum of final charges on C₂ and C₃ is :

The equivalent capacitance between $$A$$ and $$B$$ in the circuit given below, is :

A combination of parallel plate capacitors is maintained at a certain potential difference.



When a 3 mm thick slab is introduced between all the plates, in order to maintain the same potential difference, the distance between the plates is increased by 2.4 mm. Find the dielectric constant of the slab.