1.

Two identical capacitors with identical dielectric slabs in between them are connected in series as shown in. Now, the slab of one capacitor is pulled out slowly with the help of an external force F at steady state as shown. Mark the correct statement(s). .

Answer»

During the spring the process, charge (positive) flows from b to a.
During the process, the charge of capacitor B is from b EQUAL to the charge on A at all instants.
Work done by `F` is positive.
During the process, the battery has been charged.

Solution :As the dielectric slab is pulled out, then equivalent capacity of the system decrease, and hence charge supplied by the battery decreases as POTENTIAL of the system remains constant. It means charging of battery takes place and a positive charge flows from `a to b`, As the two capacitors are connected in series, so charge on bothe capacitors remains the same at all instants. from ENERGY conservation law,
`U_(i)+W_("ext")=U_(f)+"work done on battery"+DeltaH`
As dielectric slab is attracted by the plates of capcitors, to pull it out, F has to perform some work, i.e. `W_("ext")(F)gt0`.


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