a parallel plate condenser is made of 36 plates separated by paper 0.01mm thick. The area of each plate is 450 cm sq. Dielectric constant of paper is 2.5. Calculate the capacitance of the condenser.

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Solution

$C a p a c i \tan c e o f o n e c a p a c i t o r = \frac{k ε_{o} A}{d} a r e a = 450 c m^{2} = 450 \times 10^{- 4} m^{2} d i s \tan c e = 0.01 m m = 0.01 \times 10^{- 3} m^{2} C = \frac{2.5 \times 8.85 \times 10^{- 12} \times 450 \times 10^{- 4}}{0.01 \times 10^{- 3}} C = 9.9 \times 10^{- 8} f a r a d a s a l l 36 c a p a c i t o r s a r e c o n n e c t e d i n s e r i e s n e t c a p a c i \tan c e \frac{1}{C_{n e t}} = \frac{36}{C} C_{n e t} = 0.02 \times 10^{- 8} f a r a d C_{n e r} = 2 μ f a r a d$

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a parallel plate condenser is made of 36 plates separated by paper 0.01mm thick. The area of each plate is 450 cm sq. Dielectric constant of paper is 2.5. Calculate the capacitance of the condenser.

Capacitance of one capacitor = kεoAdarea = 450 cm2 = 450×10-4 m2distance = 0.01 mm = 0.01 ×10-3m2C = 2.5×8.85×10-12×450×10-40.01×10-3C = 9.9×10-8 faradas all 36 capacitors are connected in series net capacitance 1Cnet = 36CCnet = 0.02 ×10-8 faradCner = 2 μ farad