# 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.

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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