Olympiad workbookNSEPCapacitance
A parallel plate capacitor of plate area A and plate separation d is charged to potential V . Then the battery is disconnected. A slab of dielectric constant k is then inserted between the plates of the capacitor so as to fill the space between the plates completely. If Q , E and W denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted) and work d
Options
- AQ = k ₀ AE
- BQ = ₀ kAV d
- CE = V kd
- DW = ₀ AV ^2 2 ~d (1- 1 k )
Correct answer
C. E = V kd
Step-by-step solution
No solution
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