Energy stored in circuit 1 is E. If capacitors in circuit 1 and circuit 2 are connected in parallel as shown, the energy stored becomes $\frac{x E}{6}$ , find x.

alok kumar singh

Charge on C₁ = CV

Charge on C₂ = 4CV

When connected in parallel

$V_{C} = \frac{5 V}{3}$

$Q'_{1} = \frac{5}{3} C V$ $Q'_{2} = \frac{10}{3} C V$

$?$ $E = \frac{1}{2} C V^{2}$

$E' = \frac{25}{18} C V^{2} + \frac{25}{9} C V^{2}$

$\frac{25}{6} C V^{2} = \frac{50 E}{6}$

x = 50

Potential due to electric dipole on axial position at distance r from dipole is proportional to (assume r >> length of dipole)

alok kumar singh

$| E | = \frac{2 k P}{r^{3}}$

$E = - \frac{d v}{d r}$ , $v \propto \frac{1}{r^{2}}$

The force between two charged particle placed in air at separation x is F₀. Both the charged particle immerged in a medium of dielectric constant K without changing separation between two charge, then net force on one of the particle is now

alok kumar singh

In air $F = \frac{1}{4 π \in_{0}} \frac{q_{1} q_{2}}{r^{2}}$

In medium $F' = \frac{1}{4 π (k \in_{0})} \frac{q_{1} q_{2}}{r^{2}}$

$F' = \frac{F_{0}}{K}$

Five metallic plates each of area A and separated from adjacent plate by a distance d as shown in figure. The capacitance of the system across A and B is: