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Explain what would happen if in the capacitor given in Question 2.8, a 3 mm thick mica sheet (of dielectric constant =6) were inserted between the plates, (a) while the voltage supply remained connected. (b) after the supply was disconnected. |
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Answer» Solution :As for mica K = 6, HENCE NEW capacitance of the capacitor `C. = KC = 6 xx 18 pF = 108 pF` (a) If voltage supply remains connected then voltage will still remain V = 100 V and charge `Q. = C.V = 108 xx 10^(-12) xx 100 xx 10^(-8) C or 10.8 NC` (b) If supply was isconnected then charge on each plate of capacitor remains at `Q = 1.8 nC = 1.8 xx 10^(-19) C` `:.` Potential difference `V. = Q/(C.) = (1.8 xx 10^(-9))/(108 xx 10^(-12)) = 16.6V.` |
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