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The left plate of the capacitor shown in...

The left plate of the capacitor shown in the figure above carries a charge `+Q` while the right plate is uncharged at `t=0`. The total charge on the right plate after closing the switch will be:

A

`Q/2 + Cepsilon`

B

`Q/2-Cepsilon`

C

`-Q/2`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B
b. Electric field between the capacitor plates:
`E=(Q+x)/(2Aepsilon_(0))+(x)/(2Aepsilon_(0))=(1)/(2Aepsilon_(0))[Q+2x]`
Potential difference `=Ed=(d)/(2Aepsilon_(0))[Q+2x]=epsilon`
`epsilon=(Q+2x)/(2C)-x=(Q)/(2)-Cepsilon`
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