A metal plate of thickness 2mm and area `36 pi cm^(2)`- is slide into a parallel plate capacitor of plate spacing 6 mm and area `36 pi cm^(2)`. The metal plate is at a distance 3 mm from one of the plates. What is the capacitance of this arrangement ? `( "Let" (1)/(4 pi in_(0))=9xx10^(9) Nm^(2)C^(-2))`

Options

A. 8 pF
B. 15 pF
C. 25 pF
D. 20 pF

Answer

C

Explanation

When metal plate of thickness 2 mm is inserted into a parallel plate capacitor, then the arrangement is shown as,

Given, thickness of plate, `t=2` mm and area, `A=36 pi cm^(2)`
The above circuit can be redrawn as,

where, `C_(1)=(epsilonA)/(d_(1))` and `C_(2)=(epsilon_(0)A)/(d_(2))`
Now, `C_(eq)=(C_(1)C_(2))/(C_(1)+C_(2))` ( in the series combination )
`rArrC_(eq)=epsilon_(0)A[(1/d_(1) xx 1/d_(2))/(1/d_(1)xx1/d_(2))]=(epsilon_(0)A)/(d_(1)+d_(2))`
`rArr C_(eq)=(1)/(4 pi xx9xx10^(9))xx(36 pi xx 10^(-4))/(3xx10^(-3)+1xx10^(-3)) ( :.(1)/(4 pi epsilon_(0))=9xx10^(9)Nm^(2)//C^(2))`
`=1/4xx10^(-10)=25xx10^(-12)F=25 pF`
So, the correct option is (c).