shows two capacitors in series, the rigid central conducting section of length `b` being movable vertically.
.
If the potentials of the upper and lower plates are `V_(1)` and `V_(2)` respectively, then potebtial of rigid segid secton is.

Options

A. `V_(1)-((V_(1)-V_(2))x)/((a-b))`
B. `V_(1)-((V_(2)-V_(1))x)/((a-b))`
C. `V_(1)-((V_(1)-V_(2))x)/((a+b))`
D. `V_(1)-((V_(1)-V_(2))x)/((a+b))`

Answer

A

Explanation

The circuit is equicalent to two capacitors in series.
.
`C_(1)=(epsilon_(0)A)/A`
`C_(2)=(epsilon_(0)A)/((a-b-x))`
`1/C_(eq)=1/(epsilon_(0)A)[a-b-x+x]=((a-b))/(epsilon_(0)A)`
We can reduce this structure in this srrucure in the following ciruit.
`(V_(1)-V)=(V_(1)-V_(2))C_(2)/(C_(1)+C_(2))`
.
`V_(1)-V=(V_(1)-V_(2))((epsilon_(0)A)/((a-b-x)))/((epsi_(0)A)/(x)+(epsi_(0)A)/(a-b-x))`
`=(V_(1)-V_(2))x/((a-b))-(V_(1)-V_(2))x/((a-b))`.