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PhysicsQ&A Library17. 1/4 points| Previous Answers SerCP11 16.5.P.033.MI.points 1Previous Answers SercP11 16.5 P.033 MI..My Notes Ask Your TeacherAn air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm, separated by a distance of 1.90 mm. If a 15.2-V potential difference is applied to these platescalculate the following.(a) the electric field between the platesmagnitudedirection from the positive plate to the negative platekV/m(b) the capacitancePF(c) the charge on each platepCQuestion

Asked Feb 14, 2019

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Step 1

Given: Area(A) = 7.60 cm^{2} = 0.00076m^{2}

distance(d) = 1.90mm = 0.0019m

Potential difference = 15.2V

Find: (a) Electric field(E) = ?

(b) the capacitance(C) = ?

(c) the charge on each plate(Q) = ?

Step 2

(a) to calculate electric field, use the formula as follows:

Step 3

(b) to calculate capacitance, us...

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