The electric field between two parallel plates connected to a 45V battery is 1900 V/m. How far apart are the plates? Follow this format-
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- 3. PHYS101Propiem #21s Tor Questions #3 Problem #2: A 1000 μF capacitor is charged so that its stores 2.0 J of energy. Calculate the potential difference to which it has been charged. Question #3: What is the potential difference? 59 V O 71 V 82 V 63 V 79 V2. A parallel-plate capacitor has a plate area of 0.30 m² and a plate separation of 0.10 mm with only air in between. If the charge on each plate has a magnitude of 5.0 x 10-6 C, show that the energy density in its electric field is 16 Joule/m3 and the potential difference across its plates is almost 190 volts!
- To operate a given flash lamp requires a charge of 38 mC. Whatcapacitance is needed to store this much charge in a capacitor witha potential difference between its plates of 9.0 V?A 30-V battery is connected between twd parallel plates. The separation between the plates is d to be uniform, the magnitude of the electric field between the plates is equal to 0.10 cm. Assuming the electric field between the plates %3D a. 3 x 102 V/m Ob. 3 x 103 V/m Oc. 30 V/m O d. 3 x 104 V/mFind the maximum potential difference between two parallel conducting plates separated by 0.5 cm of air, given the maximum sustainable electric field strength in air to be 3.0 x 106 V/m. Hint The maximum potential difference between them is Question Help: Message instructor Submit Question V.