• # question_answer Two identical capacitors 1 and 2 are connected in series to a battery as shown in the figure. Capacitor 2 contains a dielectric slab of dielectric constant K as shown. ${{Q}_{1}}$and ${{Q}_{2}}$are the charges stored in-the capacitors. Now the dielectric slab is removed and the corresponding charges are $Q_{1}^{'}$and$Q_{2}^{'}.$Then: A)  $\frac{Q_{1}^{'}}{{{Q}_{1}}}=\frac{K+1}{K}$                    B)  $\frac{Q_{2}^{'}}{{{Q}_{2}}}=\frac{K+1}{2}$C)  $\frac{Q_{2}^{'}}{{{Q}_{2}}}=\frac{K+1}{2K}$                  D)  $\frac{Q_{1}^{'}}{{{Q}_{1}}}=\frac{K}{2}$

Solution :

Let C be the capacity of each condenser without slab. $\therefore$ When slab from capacitor 2 is removed, net capacity $=C/2\therefore \,Q_{2}^{'}=Q_{2}^{'}=\frac{{{C}_{\varepsilon }}}{2}$ Before the slab is removed${{C}_{1}}=C$ and ${{C}_{2}}=KC$             ${{C}_{Net}}=\frac{KC}{K+1}$             ${{Q}_{1}}={{Q}_{2}}=\frac{K{{C}_{\varepsilon }}}{K+1}$             $\therefore$      $\frac{Q_{2}^{'}}{{{Q}_{2}}}=\frac{{{C}_{\varepsilon }}(K+1)}{2(K{{C}_{\varepsilon }})}=\frac{K+1}{2K}.$ Hence, the correction option is [c].

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