Two 10.0-cm-diameter electrodes 0.59 cm apart form a parallel-plate capacitor. T

Question

Two 10.0-cm-diameter electrodes 0.59 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 15 V battery. What is the charge on each electrode while the capacitor is attached to the battery? Enter your answers numerically separated by a comma. What is the electric field strength inside the capacitor while the capacitor is attached to the battery? Express your answer with the appropriate units. What is the potential difference between the electrodes while the capacitor is attached to the battery? Express your answer with the appropriate units.

Explanation / Answer

d = 0.59 cm

= 0.59*10^-2 m

delta_V = 15 volts (potential diffrence across battery)

B) we know, electric field between plates of the capacitor,

E = delat_V/d

= 15/(0.59*10^-2)

= 2542 N/C

C) delta_V = 15 V

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