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Shows two capacitors in series, the rigi...

Shows two capacitors in series, the rigid central conducting section of length `b` being movable vertically.
.
The equivalent capacitance of the given structure is .

A

`(epsilon_(0)A)/((a-b-x))`

B

`(epsilon_(0)A)/((a-b))`

C

`(epsilon_(0)A)/((b-x))`

D

`(epsilon_(0)A)/((a-x))`

Text Solution

Verified by Experts

The correct Answer is:
B
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))`.
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