A parallel-plate capacitor has a capacitance of 100 pF, a plate area of 110 cm2,

Question

A parallel-plate capacitor has a capacitance of 100 pF, a plate area of 110 cm2, and a mica dielectric ( = 5.4). At 60 V potential difference, calculate the following values. (a) E in the mica V/m (b) the magnitude of the free charge on the plates C (c) the magnitude of the induced surface charge on the mica C

In the figure, the battery potential difference is 6.00 V, and each of the seven capacitors has capacitance 12 F.

(a) What is the charge on capacitor 1?
C

(b) What is the charge on capacitor 2?
C

A parallel-plate capacitor has a capacitance of 100 pF, a plate area of 110 cm2, and a mica dielectric ( = 5.4). At 60 V potential difference, calculate the following values. (a) E in the mica V/m (b) the magnitude of the free charge on the plates C (c) the magnitude of the induced surface charge on the mica C In the figure, the battery potential difference is 6.00 V, and each of the seven capacitors has capacitance 12 ½F. (a) What is the charge on capacitor 1? ½C (b) What is the charge on capacitor 2? ½C

Explanation / Answer

a) C = k e0 A/d

d = k e0 A/C = 5.4*8.85E-12*110.0E-4/100.0E-12= 5.26E-3 m

E = V/d = 60/5.26E-3=1.14E4 V/m

b) Q = C V = 100.0E-12*60 = 6.0E-9 C

c) 6.0E-9 C

2) a) C1 is in parallel with V

so V1 = 6

Q = C V = 6*12 = 72 uC

b) combine the two in series

C = 6 uF

add with one in parallel

6 + 12 = 18 uF

add with one in series

1/12 + 1/18 = 3/36 + 2/36 = 5/36

so C = 36/5

Q = 36*6/5 = 43.2 uC
things in series share charge so

43.2 uC again 18uF

so V = Q/C = 43.2/18= 2.4 V

things in parallel share voltage so voltage across 6 uF is 2.4 V

Q = 6*2.4 = 14.4 uC

things in series share charge so answer is 14.4 uC

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