1.(a) Figure 8.6 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 0.15A.

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

Answers (1)

Radius of the discs(r) = 12cm

Area of the discs(A) = \pi r^{2}=\pi (0.12)^{2}=.045m^{2}

Permittivity, \epsilon _{0}=8.85\times 10^{-12}C^{2}N^{-1}m^{2}

Distance between the two discs = 5cm=0.05m

Capacitance= \frac{\varepsilon _{0}A}{d}

=\frac{8.85\times 10^{-12}C^{2}N^{-1}m^{2}\times .045m^{2}}{0.05m}

=8.003\times 10^{-12}F

=8.003 pF

 Rate\ of \ change\ of \ potential =\frac{dv}{dt}

But

V =\frac{Q}{C}

Therefore,

Rate\ of \ change\ of \ potential=\frac{d(\frac{Q}{C})}{dt}=\frac{dq}{Cdt}=\frac{i}{C} 

=\frac{0.15A}{8.003pF}

=1.87\times 10^{10}Vs^{-1}

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