question_answer

Two identical parallel plate capacitors are connected in series to a battery of 100\[V\]. A dielectric slab of dielectric constant 4.0 is inserted between the plates of second capacitor.  The potential difference across the capacitors will now be respectively                                                [MP PMT 1992]

A)                    50 V, 50 V                       

B)            80 V, 20 V

C)                    20 V, 80 V                       

D)            75 V, 25 V

Correct Answer: B

Solution :

           \[{{C}_{eq}}=\frac{C\times 4C}{(C+4C)}=\frac{4C}{5}\] \[Q={{C}_{eq}}.V=\frac{4C}{5}\times 100=80C\] Hence \[{{V}_{1}}=\frac{Q}{{{C}_{1}}}=\frac{80C}{{{C}_{1}}}=80V\]  and \[{{V}_{2}}=\frac{80C}{4C}=20V\]