Two 10-cm-diameter electrodes 0.40 cm apart form a parallel-plate capacitor. The

Question

Two 10-cm-diameter electrodes 0.40 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 14 Vbattery.

Part A:

What are the charge on each electrode, (Q) the electric field strength inside the capacitor(E), and the potential difference between the electrodes (Delta V) while the capacitor is attached to the battery?

Part B: What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes after insulating handles are used to pull the electrodes away from each other until they are 0.80 cm apart? The electrodes remain connected to the battery during this process. (Looking for Q, E & Delta V again)

Explanation / Answer

two electrodes having the diameter is 10.0 cm

radius r = 5.0 cm

seperated by a distance d = 0.40 cm

Area of the parallel plate capacitor is

A = r^2 = ( 5.0 *10^-2 m)^2 = 0.785 *10^-2 m^2

Capacitance of the parallel plate capacitor

C = _0 A / d = 8.85 *10^-12 * 0.785*10^-2 / 0.40*10^-2 = 17.36*10^-12 F

charge of the capacitor is   Q = CV = 17.36 *10^-12 F * 19.0V = 0.243*10^-9 C

Electric field is E = V / d = 14 V / 0.40 *10^-2 m = 3.5*10^3 V /m

potential difference is V = 14.0 V - 0V = 14 V

charge on the capacitor is

Q = C V = _0 A / d * V

      = 8.85 *10^-12 * 0.785*10^-2 / 0.8 *10^-2 m * 14 V

= 121.57*10^-12 C

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