1.

The answer to each of the following questions is a single-digit integer ranging from 0 to 9. Darken the correct digit. A uniform electric field exists in between the plates of a capacitor and potential difference between the plates is found to be V_(1). Now a dielectric slab of dielectric constant 2 is inserted in between the paltes. Thickness of the salb is half of the separationbetween the plates. The potential difference between the plates is V_(2) now. Calculate 3V_(1)//V_(2).

Answer»


Solution :Let E be the electric field INTENSITY in between the PLATES and d be the separation between the plates. We can write the following:
`V_(1)=Ed`
Now when dielectric slab is introduced in between the plates then potential difference can be written as follows:
`V_(2)=E(d-t)+(E//K)(t)`
Here t is `d//2` and K = 2. On substituting we GET the following:
`V_(2)=3V_(1)//4` hence `3V_(1)//V_(2)=4`.


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