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29 Jul 2009 19:04:58 IST
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A capacitor is connected to a battery. a dielectric is then inserted. (battery is not removed ).What is the effect on electric field?Explain with reasons.
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29 Jul 2009 20:42:25 IST
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The electric field decreases by a factor K(dielectric constan), due to induction of charges on the dielectric.
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30 Jul 2009 12:46:35 IST
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T he potential difference V between the plates of acapacitor is reduced to V/K when a dielectric is introduced. Because the potentialdifference between the plates equals the product of the electric field and theseparation d, the electric field is also reduced. Thus, if E0 is the electric field withoutthe dielectric, the field in the presence of a dielectric isE=E0/K
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30 Jul 2009 16:49:11 IST
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HEY TUSHAR U R WRONG DUDE.. HOW CAN THE POTENTIAL DIFFERNCE CHANGE WHEN THE BATTERY IS STILL CONNECTED ?? THE POTENTIAL DIFFERNCE WILL REMAINS SAME BUT THE CHARGE ON CAPACITOR WILL CHANGE AND THE ELECTRIC FIELD WILL BECOME E / K AS TRUELY SAID BY ANUBHAV
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30 Jul 2009 18:54:02 IST
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hello dear iam writing the concept of this question:: when the battery is attached to the capacitor than voltage is constant across the capacitor,there is no change of the voltage across the capacitor note:: u can solve this question by the given hint
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30 Jul 2009 22:06:58 IST
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So EF would'nt change
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30 Jul 2009 22:23:35 IST
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arrey yaar see here the volatge acroos the capacotior is remaining constant so electric field is remaining constant but u might expect it to reduce actually this effect is cancelled by the charge tat the battery delivers to the plates(adiitional) so charge increases k times and capcitance too increases for better understanding refer resnick halliday krane cheers
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31 Jul 2009 09:05:29 IST
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Sorr i meant electric field remains constant
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31 Jul 2009 16:09:18 IST
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look......when there is no dielectric then suppose the electric field is from right to left (assuming right plate to be positive and left to be negative)........wen u insert the dielectric, induction will take place and right plate of the dielctric will become negative and left positive so dere will be electric field in the opposite direction,,so the net electric field will be reduced.........due to opposite directions.....o.k
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31 Jul 2009 17:48:00 IST
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No .......... well when a battery remains connected the potential V remains a constant . the distance distance between the plates is not changing....... so electric field V/D must remain a constant. Here the effect of polarisation is taken into account as seen in the case of isolated capacitor in NCERT.Hence the EF must remain a constant
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1 Aug 2009 12:49:46 IST
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sorry dear but it will nt be constant ...........sure sure
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1 Aug 2009 12:51:01 IST
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yeah ! electric field will change as the charge on capacitor is changing
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1 Aug 2009 21:00:37 IST
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exactly yaar it has to change and dat too towards the lower part of it..........it will decrease.....cm along with ur logics if u have any!!!!!!!!!!!!
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