A parallel plate capacitor has a plate separation of 2.0 mm and a plate area of

Question

A parallel plate capacitor has a plate separation of 2.0 mm and a plate area of 550 mm2 . The space between the plates is filled by Teflon (see Table 23.1 of your text). The voltage between the plates is 6.8 kV.

(a) {1} What is the capacitance?

(b) {1} Calculate the charge on the capacitor.

(c) {1} How many electrons would this represent (on the negatively charged plate)?

(d) {1} What is the strength (magnitude) of the electric field inside the capacitor?

(e) {1} Would this exceed the dielectric breakdown field?

(f) {1} Calculate the energy stored in the capacitor.

Explanation / Answer

a) C = e0 A /d
C = 8.854 x 10-12 x (550 x 10-6) m^2 / 2 x 10-3 m
C = 2.43 x 10-18 F

b) Q = CV
Q = 2.43 x10-18 x 6.8 x 10^3   = 1.656 x 10-14 C

c) Q = ne
n = 1.656 x 10-14 / (1.6x 10-19 )   = 103481.125 electrons

d)
E.d = V
E = 6.8 x 10^3 / 2x10-3   = 3.4 x 10^6 N/C

a) C = e0 A /d
C = 8.854 x 10-12 x (550 x 10-6) m^2 / 2 x 10-3 m
C = 2.43 x 10-18 F

b) Q = CV
Q = 2.43 x10-18 x 6.8 x 10^3   = 1.656 x 10-14 C

c) Q = ne
n = 1.656 x 10-14 / (1.6x 10-19 )   = 103481.125 electrons

d)
E.d = V
E = 6.8 x 10^3 / 2x10-3   = 3.4 x 10^6 N/C

e) ..........information insuffecient

f) E = CV^2 / 2
         = 2.43 x 10-18 x (6.8 x 10^3 )^2 /2   = 5.62 x 10 - 11 J

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