1. Two 10.0-cm-diameter electrodes 0.5 cm apart form a parallel-plate capacitor.

Question

1. Two 10.0-cm-diameter electrodes 0.5 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 15 V battery. Calculate the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes: a. While the capacitor is attached to the battery. b. After insulating handles are used to pull the electrodes away from each other until they are 1.0 cm apart. Note: the electrodes remain connected to the battery during this process. c. After the original electrodes (not the modified electrodes of part b) are expanded until they are 20.0 cm in diameter while remaining connected to the battery.

I have no idea where to start :(

Explanation / Answer

Given: diameter of the plates d = 10 cm radius of the plates r = 5 cm                                  = 5*10-2 m                                  = 0.05 m distance between the plates d = 0.5 cm                                                = 0.005 m potential difference between the plates V = 15 V ............................................................................................. Here it form a parallel plate capacitor capacitance of a parallel plate capacitor C = 0A/d                                                                      = (8.85*10-12)(3.14)( 0.05 m )/ 0.005 m                                                                      = 2.7*10-10 F charge on each plate of capacitor Q = CV                                                          = ( 2.7*10-10 )(15 V)                                                          = 3.7*10-9 C Electric field strength E = V/d                                        = 15 / 0.005                                        = 3*103 N/C .................................................................................................................... d = 1 cm    = 0.01 m capacitance of a parallel plate capacitor C = 0A/d                                                                      = (8.85*10-12)(3.14)( 0.05 m )/ 0.01 m                                                                      = 1.3*10-10 F charge on each plate of capacitor Q = CV                                                          = ( 1.3*10-10 )(15 V)                                                          = 1.9*10-9 C Electric field strength E = V/d                                        = 15 / 0.01                                        = 1500 N/C capacitance of a parallel plate capacitor C = 0A/d                                                                      = (8.85*10-12)(3.14)( 0.05 m )/ 0.01 m                                                                      = 1.3*10-10 F charge on each plate of capacitor Q = CV                                                          = ( 1.3*10-10 )(15 V)                                                          = 1.9*10-9 C Electric field strength E = V/d                                        = 15 / 0.01                                        = 1500 N/C

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