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![[Post New]](/templates/default/images/icon_minipost_new.gif) 9 Dec 2007 20:37:19 IST
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A capacitor stores a charge of 50  .when the gaP between the plates is filled with a dielictric a charge of 100 *10 to the power-6 charge flows through the battery. find the dielictric constant . pls explain in detail
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11 Dec 2007 18:11:06 IST
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Let C(k) b d capacitance wen d dielectric slab is inserted between d 2 plates of d capacitor....... n C b d capacitance widout ne dielectric slab.... den,
dielectric constant ,k= C(k)/ C
= q(k)/q
=100x 10^-6/ 50x 10^-6
=2
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11 Dec 2007 18:40:53 IST
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no K=3. if correct ill post method.
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11 Dec 2007 19:20:21 IST
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pannaguma......... plz do tel me hwz k=3......... plz na do nudge me..........
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11 Dec 2007 20:48:01 IST
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See,when Capacitor is filled with dielectric,capacitance becomes k times
Initial charge = 50uC,Capacitance=C(say)
Final charge = 100uC+50uC=150uC,Final capacitance = kC
Since battery is still connected V will not change,therefore q/C=constant
50/C = 150/kC
Therefore k = 3
Mistake ur doing is that an ADDITIONAL 100uC is flowing in in addition to the 50uC already present
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12 Dec 2007 15:08:42 IST
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yeah mystic's expl seems right.
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