In the following circuit, the switch S is closed at t = 0. The charge on the capacitor C₁ as a function of time will be given by $$\left( {{C_{eq}} = {{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right)$$

A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K = 5/3 is inserted between the plates, the magnitude of the induced charge will be :

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 :