Find the equivalent capacitance between the point A and B in figure .

Explanation

Let us connect a battery between the points A and B. The charge distribution is shown jin figure. Suppose the positive terminal of the battery supplies a charge +Q and the negative terminal a charge -Q. The charge Q is divided between plates a can e. A charge `
`Q_1`
` goes to the plate a and the rest `
`Q-Q_1`
` goes to the plate e. The charge -Q supplied by the nagative terminal is divided between plates d and h. Using the symmetry of the figure, charge `
`-Q_1`
` goes to the plate h and `
`-(Q - Q_1)`
` to the plate d. This is because if you look into the circuit from A or from B, the circuit looks identical. The division of charge at A and at B should, therfore, be similar. The charges on the other plates may be written easily. The charge on the plate l is `
`2 Q_1 - Q`
` which ensures that the total charge on plates b,c and i remains zero as these three plates form an isolated system. we have `
`V_A V_B =(V_A - V_D) + (V_D - V_B)`
` = `
`(Q_1)/(C_1) + (Q- Q_1)/(C_2)`
` ... (i) Also, `
`V_A - V_B = (V_A - V_D) + (V_D - V_E) + (V_E - V_B)`
` = (Q_1)/(C_1) + (2Q_1 - Q)/(C_3) = (Q_1)/(C_1)`
`. ....(ii) We have to eliminate `
`Q_1`
` from these equations ot get the equivalent capacitance `
`Q/ (V_A- V_B)`
`. The first equation may be written as `
`V_A - V_B = Q_1 ((1)/(C_1) - (1)/(C_2)) + (Q)/(C_2)`
` or, `
`(C_1 C_2)/(C_2 - C_1) (V_A - V_B ) = Q_1 + (C_1)/(C_2 - C_1) Q`
`. The second equation may be written as `
` V_A - V_B = 2Q_1((1)/(C_1)+(1)/(C_3)) - (Q)/(C_3) or, (C_1 C_3)/(2(C1 + C_3) (V_A - V_B) = Q_1 - (C_1)/(2(C_1 + C_3) Q. ...(IV) Subtracting (iv) from (iii) `
`(V_A - V_B)[(C_1 C_2)/(C_2 - C_1) - (C_1 C_3) /(2(C_1 + C_3))]`
` `
` = [(C_1)/(C_2 -C_1) +(C_1)/(2(C_1 + C_3)] Q`
` or `
`(V_A - V_B) [2 C_1 C_2(C_1 + C_3) + (C_2 - C_1)]`
` `
`=C_1[2(C_1 + C_3) + (C_2 - C_1)] Q`
` or, `
` C = (Q)/(V_A - V_B) = (2 C_1 C_2 + C_2 C_3 + C_3 C_1)/( C_1 + C_2 + 2C_3)`
`.