In a circular parallel plate capacitor radius of each plate is 5 cm and they are separated by a distance of 2mm. Calculate the capacitance and the energy stored. When it is charged by connectig battery of 200 V. (given `epsilon_(0) = 8.854 xx 10^(-12) Fm^(-1)`)

Explanation

Given : `d=2xx10^(-3)` m,
`r=5xx10^(-2)`m ,
V=200 V ,
C=? , E= ?
Area of circular plate , `A=pir^2`
i.e., `A=3.14xx(5xx10^(-2))^2`
`A=78.55xx10^(-4) m^2`
We know that ,
`C=(epsilon_0A)/d=(8.854xx10^(-12)xx78.55xx10^(-4))/(2xx10^(-3)) F`
i.e., `C=34.774xx10^(-12) ` F
Also , energy stored in a capacitor
`=1/2CV^2`
`=1/2xx34.774xx10^(-12)xx(200)^2`
`=6.955xx10^(-7)` J
Hence Energy stored = `6.955xx10^(-7) ` J