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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5.33 volt

6.34 volt

4.33 volt

7 volt

answer is C.

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Detailed Solution

$c =$ Initial capacitance of each capacitor.

$∴ C_{e} = \frac{c \times c}{c + c} = c / 2 ∴ Q =$ charge on each capacitor $C . E = \frac{C E}{2}$

$∴ \frac{Q}{C} = \frac{1}{C} \times \frac{C E}{2} = 4 \Rightarrow E = 8 volt$

Finally, $c_{2}^{1} = 2 c, c_{e}^{1} = \frac{c \times 2 c}{c + 2 c} = \frac{2 c}{3}, Q^{1} = c_{e}^{1} E = \frac{2 C E}{3}$

$∴$ Final p.d. across $1 = Q^{1} / C = \frac{2 E}{3} = \frac{2 \times 8}{3} volt=5.33 volt$

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