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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 difference will then be `V = E_(0)(1/4d) + 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)/4K while thefree CHARGE `Q_(0)`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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