Three identical capacitors are combined differently. For the same voltage to each combination, the one that stores the greatest energy is

$A$ $2$ $\mu F$ capacitor is charged to a potential $=$ $10\,V$. Another $4$ $\mu F$ capacitor is charged to a potential $=$ $20\,V$. The two capacitors are then connected in a single loop, with the positive plate of one connected with negative plate of the other. What heat is evolved in the circuit?……$\mu J$

The energy stored in the electric field produced by a metal sphere is $4.5\, J$. lf the sphere contains $4\,\mu C$ charge, its radius will be…….$mm$ : [Take : $\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}\,N – {m^2}\,/{C^2}\, ]$

The plates of a parallel plate capacitor have an area of $90 \,cm ^{2}$ each and are separated by $2.5\; mm .$ The capacitor is charged by connecting it to a $400\; V$ supply.

$(a)$ How much electrostatic energy is stored by the capacitor?

$(b)$ View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume $u$. Hence arrive at a relation between $u$ and the magnitude of electric field $E$ between the plates.

A capacitor of capacitance $50 \; pF$ is charged by $100 \; V$ source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is $\dots \; nJ$.