17. 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 platepC

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Asked Feb 14, 2019
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17. 1/4 points| Previous Answers SerCP11 16.5.P.033.MI.
points 1Previous Answers SercP11 16.5 P.033 MI..
My Notes Ask Your Teacher
An 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 plates
calculate the following.
(a) the electric field between the plates
magnitude
direction from the positive plate to the negative plate
kV/m
(b) the capacitance
PF
(c) the charge on each plate
pC
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17. 1/4 points| Previous Answers SerCP11 16.5.P.033.MI. points 1Previous Answers SercP11 16.5 P.033 MI.. My Notes Ask Your Teacher An 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 plates calculate the following. (a) the electric field between the plates magnitude direction from the positive plate to the negative plate kV/m (b) the capacitance PF (c) the charge on each plate pC

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Step 1

Given:      Area(A) = 7.60 cm2 = 0.00076m2

                   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:

 

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Step 3

(b) to calculate capacitance, us...

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Electric Charges and Fields

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