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Capacitors and capacitance

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By Expert Faculty of Sri Chaitanya
Question

Figure shows a capacitor made of two circular plates, each of radius 12 cm and separated by 5.0 mm. The capacitor is being charged by an external source (not shown in figure). The charging current is constant and equal to 0.15 A. Calculate the rate of change of potential difference between the plates (in 109 V/s).

Moderate
Solution

Area of one of the plates, A=π12×1022 m2

Distance between the plates, d=5.0mm=5×103 m

Capacitance of the capacitor, C=ϵ0Ad

or C=14π×9×109×π12×10225×103

    =80×1012 F=80 pF

Charging current, I=dQdt=ddt(CV)

 or I=CdVdt or dVdt=IC

Rate of change of p.d.

=dVdt=IC=0.15 A80×1012 F=1.87×109 V/s


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