Two 10.0-cm-diameter electrodes 0.50 cm apart form a parallel-plate capacitor. T
ID: 1536022 • Letter: T
Question
Two 10.0-cm-diameter electrodes 0.50 cm apart form a parallel-plate capacitor. The electrodes are connected to the terminals of a 15 V battery. What are (i)the electric field inside the capacitor, (ii) the charge on each electrode, and (iii) the potential difference between the electrodes: (a) while the capacitor is attached to the battery and (b) After insulating handles are used to pull the electrodes from each other until they are 1.0 cm apart? The electrodes remain connected to the battery during this process.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 )2/ 0.005 m
= 1.39*10-11 F
charge on each plate of capacitor Q = CV
= ( 1.39*10-11 )(15 V)
= 2.09*10-10 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 )2/ 0.01 m
= 6.95*10-12 F
charge on each plate of capacitor Q = CV
= ( 6.95*10-12 )(15 V)
= 1.04*10-10 C
Electric field strength E = V/d
= 15 / 0.01
= 1500 N/C
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