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The area of each plate of a parallel pla...

The area of each plate of a parallel plate capacitor is A = `600 cm^2` and their separation is d = 2.0 mm. The capacitor is connected to a 200 V dc source. Find out
the surface density of charge on any plate. Given, `in_0 = 8.85xx10^(-12)F*m^(-1)`.

Text Solution

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Here, `A= 600 cm^2 = 600xx 10^(-4) m^2 = 6xx10^(-2)m^2 , d= 2.0 mm= 2xx10^(-3)m`.
If `sigma`= surface density of charge on may plate, then
`E = (sigma)/(in_0)`
`:. sigma = in_0E = (8.85xx10^(-12))xx10^(5)`
`= 8.85 xx 10^(-7)C*m^(-2)`.
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