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

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

Explanation

Capacitance of parallel plate capacitor is,
`C=(in_(0)A)/(d)`
`=(in_(0)(pi R^(2)))/(d)`
`=((8.85xx10^(-12))(3.14)(0.12)^(2))/((0.005))`
`therefore C=80xx10^(-12)F=80 pF` (picofarad)
According to formula capacitance of capacitor,
`C=(Q)/(V)`
`therefore Q=CV`
`therefore (dQ)/(dt)=C(dV)/(dt)`
`therefore I=C(dV)/(dt)`
`therefore (dV)/(dt)=(I)/(C )`
`therefore (dV)/(dt)=(0.15)/(80xx10^(-12))`
`therefore (dV)/(dt)=1.875xx10^(9)Vs^(-1)`