Q.

Two identical parallel plate capacitor are connected in series combination as shown in figure. Potential difference across capacitor1 is found to be 4 volt. Now separation between the plates of capacitor 2 is halved. Then new potential difference across 1 will be

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a

4.33 volt

b

6.34 volt

c

5.33 volt

d

7 volt

answer is C.

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Detailed Solution

c=Initial  capacitance of each capacitor. ∴Ce=c×cc+c=c/2 ∴ Q= charge on each capacitor C.E=CE2∴QC=1C×CE2=4⇒E=8 voltFinally, c21=2c,   ce1=c×2cc+2c=2c3,   Q1=ce1E=2CE3∴ Final p.d. across 1=Q1/C=2E3=2×83volt=5.33 volt
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Two identical parallel plate capacitor are connected in series combination as shown in figure. Potential difference across capacitor1 is found to be 4 volt. Now separation between the plates of capacitor 2 is halved. Then new potential difference across 1 will be