Electric potential and capacitance Questions

The applied potential difference in the circuit shown is V = 10 sin 100 t where V is in volt and t is in second. If the power factor of the circuit is 1/√2, then the value of capacitance C of the circuit, nearly is

The charge on capacitor of capacitance 15μF in the figure given below is:

A parallel plate capacitor consists of two circular plates each of radius 2 cm, separated by a distance of 0.1 mm. If voltage across the plates is varying at the rate of 5 × 10¹³ V/s, then the value of displacement current is :

Two capacitors of capacitances 3μF and 6μF are charged to a potential of 12V each. They are now connected to each other, with the positive plate of each joined to the negative plate of the other. The potential difference across each will be-

Two point charges −4μC and 4μC , constituting an electric dipole, are placed at (− 9, 0, 0) cm and (9, 0, 0) cm in a uniform electric field of strength 104 NC−1 . The work done on the dipole in rotating it from the equilibrium through 180° is :