there is a five plate capacitor with area of plates 34meter square and distance between each plate is .2m?

Question

how to find equivalent capactance?................................... will we take 1st positively charged 2nd -vely charged ,3rd positevily charged andso on or.................will there charges induced on every plate both +ve and -ve such as.................
I+ -I+ -I+ -I+ I.............................HELP ME WITH REASON

Answer 1

if the 5 plates are connected in series the equivalent capacitance is Ce

1/Ce = 1/C1+1/C2+1/C3+1/C4

C1=C2=C3=C4= C results Ce= 1/4 C

C = e x S/d

e - epsilon - electric permittivity; for air e= 1 F/m and for vacuum e = 8.854187817 x 10^-12 F /m
S - plate area
d - distance between plates

Answer 2

I'm open to correction here, but these five plates will act like 4 identical capacitors operating in series.

capacitors in series act like resistors in parallel.

Therefore, the total capacitance, Ct, is given by

1/Ct = 1/C1 + 1/C2 + 1/C3 + 1/C4 = 4/C

(as the capacitors are identical)

Therefore Ct = C/4