Two 13.0-cm-diameter electrodes 0.50 cm apart form a parallel-plate capacitor. T

Question

Two 13.0-cm-diameter electrodes 0.50 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 14 V battery.

Part A) What is the charge on each electrode while the capacitor is attached to the battery?

Part B) What is the electric field strength inside the capacitor while the capacitor is attached to the battery?

Part C) What is the potential difference between the electrodes while the capacitor is attached to the battery?

Part D) What is the charge on each electrode after insulating handles are used to pull the electrodes away from each other until they are 1.9 cm apart? The electrodes remain connected to the battery during this process.

Part E) What is the electric field strength inside the capacitor after insulating handles are used to pull the electrodes away from each other until they are 1.9 cm apart? The electrodes remain connected to the battery during this process.

Part F) What is the potential difference between the electrodes after insulating handles are used to pull the electrodes away from each other until they are 1.9 cm apart? The electrodes remain connected to the battery during this process.

Explanation / Answer

Given circular parallel plate capacitor with electrodes of

diameter = 13 cm, radius r = 6.5 cm = 0.065 m,

separation d = 0.5 cm = 0.005 m

battery v = 14V

Part A ) charge on each electrode is Q = c*V ==> Q = (epsilon not*a/d)*V

   C = (8.854*10^-12 *pi*0.065^2/0.005) F = 2.350*10^-11 F = 23.5 pF

now the charge Q = c*V = 23.5*10^-12*14 C = 3.29*10^-10 C = 0.329 nC

Part B) we have relation is E = V/d ==> E = 14/0.005 N/C = 2800 N/C

Part C) V = E*d = 14 V

Part D)

   when the electrodes are separed by adistance d = 1.9 cm = 0.019 m , the capacitance is

C = (epsilon not*a/d)
C = (8.854*10^-12 *pi*0.065^2/0.019) F = 6.185324696*10^-12 F = 6.185325 pF

   now charge is Q = c*V = 6.185325*10^-12*14 C = 8.659455*10^-11 C

Part E) relation is E = V/d ==> E = 14/0.019 N/C = 736.84211 N/C

Part F) V = E*d = 14 V

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