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A slab of material of dielectric constant K has the same area as the plates of a parallel-platecapacitor but has a thickness ((3)/(4)) d where d is the separation of the plates. How Is the capacitance changed when the slab ls 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 vacuum (air) and here the potential difference is V 0. If the dielectric is now inserted, the electric field in the dielectric will be`E = (E_(0))/(K)`where K is dielectric constant The potential difference will then be `V= (V_(0))/(4) +(3)/(4) .(V_(0))/(K)` `= (E_(0)d)/(4) +(3)/(d) .(E_(0)d)/(K)` `=E_(0)d[(1)/(4)+(3)/(4K)]` `= V_(0)[(K+3)/(4K)]` Hence, the potential difference decreases by the factor `(K+3)/(4K)`while the free charge `Q_(0)`on the plates remains unchanged. The capacitance thus increase, `C = (Q_(0))/(V) = (4K)/(K+3) .(Q_(0))/(V_(0))` `:. C = (4K)/(K+3).C_(0)[ because (Q_(0))/(V_(0))=C_(0)]` |
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