The figure shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm . The capacitor is beging charged by an external source (not shown in the figure. ) The charing is constant and equal to 0.15 A

Calculate the capacitance and the rate of charge of potential difference between the plates.

Explanation

Given `epsi_(0)=8.85xx10^(-12) C^(2)N^(-1) m^(-2)`
Here , R=12 cm =0.12 m, d=5.0 mm=`5xx10^(-3)m `, I=0.15 A
Area , A=`pi R^(2)=3.14xx(0.12)^(2) m^(2)`
We know that capacity of a parallel plate capacitor is given by
`C=(epsi_(0)A)/(d)=(8.85xx10^(-12)xx3.14xx(0.12)^(2))/(5xx10^(-3))=80.1 xx10^(-12) F`
`=80.1 ` PF
Now `q=CV` or `(dq)/(dt)=Cxx(dv)/(dt)` or `I=C.(dv)/(dt)`
Or `(dv)/(dt)=(I)/(C)=(0.15)/(80.1xx10^(-12))=1.87xx10^(9) Vs^(-1)`