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:

Options

A. `Q/2+Cepsilon`
B. `Q/2-Cepsilon`
C. `-Q/2`
D. `-Cepsilon`

Answer

B

Explanation

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`