Question

# A parallel-plate capacitor is connected to a battery as shown in the figure. A conducting plate is inserted mid-way between the two plates. Take the plate area as A. The distance between each plate in the new configuration is d. If the middle and upper plates are short circuited, the extra charge flown through the battery is

A
ε0AVd
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B
Zero
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C
ε0AV2d
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D
2ε0AVd
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Solution

## The correct option is C ε0AV2dWe have, Ci=ε0A2d Qi=CiVi Qi=ε0A2d×V Now upper plates are short circuited, So, only one plate remains in picture and the potential difference across it is V. Cf=ε0Ad Qf=CfVf Qf=ε0Ad×V So, the extra charge flown is Qf−Qi ε0AdV−ε0AV2d =ε0AV2d

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