= 2.00 x 105 V cos [(800 rad = 2.00 x 105 V cos [(800 rad
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- (a) Determine the electric field strength between two parallel conducting plates to see if it will exceed the breakdown strength for air (3 ✕ 106 V/m). The plates are separated by 3.48 mm and a potential difference of 5515 V is applied. V/m(b) How close together can the plates be with this applied voltage without exceeding the breakdown strength? mm(22%) Problem 6: An air-filled capacitor has capacitance Co. When it is connected to a battery with EMF Vo it has charge Qo and stored energy Uo. Once the capacitor has charged, then the switch is opened, and finally a slab of material with dielectric constant K is inserted into the gap. -Vo + Vo d DV 20% Part (a) Which of the following physical quantities does not change when the dielectric material is inserted? Grade = 100% Correct Answver Student Final Submission Feedback Not available until end date The charge stored on the capacitor plates. Correct! Grade Summary Deduction for Final Submission 0% Deductions for Incorrect Submissions, Hints and Feedback [?] 0% Student Grade = 100 - 0 - 0 = 100% Submission History All Date times are displaved in Eastern Standard Time.Red submission date times indicate late work. Date Time Answer Hints Feedback 1 Feb 27, 2021 7:44 PM The charge stored on the capacitor plates. E * 20% Part (b) Enter an expression for the potential difference between…1. (a) Will the electric field strength, E, between two parallel conducting plates exceed the breakdown strength for air (3·106 V -) if the plates are separated by 2.1 mm and a potential difference of 5200 V m is applied? Determine this by calculating the electric field strength between two parallel conducting plates. V E = Pick a "Yes" or "No" (b) How close together can the plates be with this applied voltage before the air breaks down? d = mm
- [20](a)Will the electric field strength between two parallel conducting plates exceed the breakdown strength of dry air, which is 3.00 × 106 V/m , if the plates are separated by 2.00 mm and a potential difference of 5.0 × 103 V isapplied? (b) How close together can the plates be with this applied voltage?How much work W is required to move a 3.0-C positive charge from the negative terminal of a 6.0-V battery to the W = positive terminal?