Question

A parallel plate air capacitor has rectangular plates, each of area 20 cm2 separated by a distance of 2mm. the potential difference between the plates is 500 volt. Calculate (i) Its capacitance (ii) The charge on each plate (iii) The electric field intensity between the two plates.

Answer 1

Explanation:

Given A parallel plate air capacitor has rectangular plates, each of area 20 cm2 separated by a distance of 2 mm. the potential difference between the plates is 500 volt. Calculate (i) Its capacitance (ii) The charge on each plate (iii) The electric field intensity between the two plates.

• Given area = 20 cm^2 = 0.002 m^2
• d = 2 mm = 2 x 10^-3 m
• so εo = 8.85 x 10^-12 C^2 / N-m^2
• potential difference v = 500 V
• so the capacitance of the capacitor is given by
• C = Akεo / d
•   = 0.002 x 1 x 8.85 x 10^-12 / 2 x 10^-3
•   = 0.00885 x 10^-9
•   = 8.85 x 10^-12 F
• Now we need to find the charge on each plate:
• So V = Q / C
• Or Q = CV
•          = 8.85 x 10^-12 x 500
•         = 4425 x 10^-12 C
• Now the electric field intensity between the plates will be  
•        E = v / d
•      E = 500 / 2 x 10^-3
•     E = 250 x 10^3 N/C

Reference link will be

https://brainly.in/question/16267062