Two identical capacitors are connected as shown. A dielectric slab Is introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected. Compare the capacitance, the charge, potential difference and stored energy of each capacitor before and arter introducting of dielectric slab.

Explanation

`C_(1)=C_(2)=C_(0)`
`V_(1)=V_(2)=(V_(0))/(2)`
`U_(1)=U_(2)=(1)/(2)C_(0)((V_(0))/(2))^(2)=(C_(0)V_(0)^(2))/(8)`
After introduction of slab:
`C'=C_(0),C'_(2)=KC_(0)`
`V'_(1)=((KC_(0))/(KC_(0)+C_(0)))V_(0)=(KV_(0))/(K+1)` ltbr) `V'_(2)=V_(0)-V'_(1)=(V_(0))/(K+1)`
`Q'_(1)=Q'_(2)=C'_(1)V'_(1)=(KC_(0)V_(0))/(K+1)`
(In series, charge on each capacitor is same.)
`U'_(1)=(1)/(2)V'_(1)V'_(1)^(2)=(1)/(2)C_(0)((KV_(0))/(K+1))^(2)=((K)/(K+1))^(2)(C_(0)V_(0)^(2))/(2)`
`U'_(2)=(1)/(2)C'(2)V'_(2)^(2)=(1)/(2)KC_(0)((V_(0))/(K+1))^(2)=(K)/((K+1)^(2))(C_(0)V_(0)^(2))/(2)`
Comparison: `C'_(1)=C_(1),C'_(2)gtC_(2)`
`V'_(1)=(KV_(0))/(K+1)=(V_(0))/(1+(1)/(K))`
Since `K gt 1,(1+(1)/(K)) lt 2`
`V'_(1) gt V_(1),V'_(2) lt V_(2),(V_(1)+V_(2)=V'_(1)+V'_(2))`
`Q'_(1) gt Q_(1),Q'_(2) gt Q_(2)`
`U'_(1) gt U_(1),U'_(2) lt U_(2)`
You can compare by taking `K=2`.