Five plates are arranged as shown in the figure and connected across a battery of emf V. The separation between each plate is d and surface area of each plate is A.

Options

A. Equivalent capacity between A and B is `epsilon_0A//5d`.
B. Equivalent capacity between A and B is `3epsilon_0A//5d`.
C. Charge on plate 1 is `epsilon_0AV//5d`
D. Charge on plate 3 is `3epsilon_0AV//5d`.

Answer

B, C, D

Explanation

`C = (A epsilon_(0))/(d)`
`(1)/(C_(eq)) = (1)/(C ) + (2)/(3C) = (5)/(3C)`
Or `C_(eq) = (3C)/(5) = (3Aepsilon_(0))/(5d)`
Alternative method

`C_(ed) = (O)/(V) = (x + y)/(V_(AB))`
Potential of 1 and 4 is same
`(y)/(Aepsilon_(0)) = (2x)/(Aepsilon_(0))` or ` y = 2x`
`V = ((2 y + x)/(A epsilon_(0))) d, C_(eq) = ((x + 2x)A epsilon_(0))/((5x)d) = (3Aepsilon_(0))/(5d)`