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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

`-Cepsilon`

Text Solution

Verified by Experts

The correct Answer is:
B
Electric field between the capacitor plates
`(sigma_(1))/(2epsilon_(0))+((-sigma_(2)))/(2epsilon_(0))`
`E=(Q+x)/(2Aepsilon_(0))+x/(2Aepsilon_(0))=1/(2Aepsilon_(0))[Q+2x]`
`implies` Potential different `Ed=d/(2Asepsilon_(0))[Q+2x]=epsilon`
`impliesepsilon=(Q+2x)/(2C)=-x=Q/2-Cepsilon`
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