IIT JEE , AIEEE Forum - Physics , Electricity

striving4perfection

Hello

confusions have a very strong grip on me 'coz I always end up in some or the other confusions........For example , today i got stuck up wid this problem....

Two identical capacitors are connected in series. A dielectric slab is inserted between the plates of 1 capacitor, the battery remaing connected so that a constant potential difference V is maintained.Describe qualitatively wat happens to the charge, the capacitance, the potential difference for each capacitor.

Take capacitance of each capacitor to be C.

WOULD DEFINITELY RATE U.

striving4perfection

Plzzzzzzzz..... reply . Jot down anything that comes to ur mind....

tarun007

hi miss striving4perfection

practice practice and more practice is the only way out of confusion
regarding this q:

at first: both of them have equal capacitance c so the circuit capacitance wud be c/2 hence charge on each capacitor wud be cv/2
potential diff across each capacitor wud be v/2

when u insert a slab then one of the capacitor will have capacitance kc
and the circuit capacitance will be kc/(k+1)
hence charge on each capacitor wud be kcv/(k+1)
potential diff on slab capacitor=charge/capacitance=v/(k+1)
and on the other 1=kv/(k+1)

iitkgp_bipin

There is nothing to worry, practice is the only way.
Just go step by step.

When tha capacitors are connected in series, effective capacitance = C/2
Net voltage = V
Net charge = CV/2
Since they are in series charges on both of them = CV/2
Potential difference across both of them = V/2

When a slab of dielectric constant k is inserted in one of them.
Their capacitances become C and kC.
Net capacitance in series = (kC)(C) / (kC+C) = kC/(k+1)
Net volatge remains same = V
Charge on each of the them = kCV/(k+1)
Potential difference across C = charge/C = kV/(k+1)
Potential difference across kC = charge/kC = V/(k+1)

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