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InterviewSolution
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A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness (3/4)d. where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates? |
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Answer» Solution :Let `E_(0)=V_(0)/d` be the electric field between the PLATES when there is no dielectric and the potential difference is `V_(0)`. If the dielectric is now inserted, the electric field in the dielectric will be `E=E_(0)/K`. The potential will then be `V=E_(0) (1/4 d)+E_(0)/K (3/4 d)=E_(0) d(1/4+3/(4K))=V_(0) (K+3)/(4K)` The potential difference decreases by the factor `((K+3))/(K)` while the free charge `Q_(a)` on the plates remains unchanged. The capacitance thus INCREASES. `C=Q_(0)/V=(4K)/(K+3) (Q_(0))/(V_(0))=(4K)/(K+3) C_(0)` |
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