A certain series `RC` circuit is formed using a resistance `R`, a capacitor without dieiectric having a capacitance `C=2F` and a battery of emf `E=3V`. The circuit is completed and it is allowed to attain the steady state. After this, at `t=0`, half the thickness of the capacitor is filed with a dielectric of constant `K=2` as shown in the figure. The system is again allowed to attain a steady state. What will be the heat generated (in joule) in the capacitor between `t=0` and `t=oo` ?

Explanation

Initial charge on capacitor `=CE`
Initial potential energy of capacitor `=CE^(2)//2`
Now `C=epsilon_(0)A//d`
And new capacitance: `C'=(epsilon_(0)A)/(d//2)`, in series with `(epsilon_(0)kA)/(d//2)`
`rArr C'=(2epsilon_(0)A)/(d)` series `(4epsilon_(0)A)/(d)`
`=(epsilon_(0)A)/(d)((2xx4)/(2+4))=(4epsilon_(0)A)/(3d)=(4)/(3)C`
`rArr` New charge `=(4)/(3)CE`
New energy `=(1)/(2)xx(4)/(3)CE^(2)=(2)/(3)CE^(2)`
Now `W_("but")=DeltaH+DeltaU`
`rArr E((4)/(3)CE-CE)=DeltaH+(2)/(3)CE^(2)-(1)/(2)CE^(2)`
`rArr DeltaH=(1)/(6)CE^(2)`