1 picofarad = 1 pF = 1×10 -12 F. A parallel-plate capacitor, of area 6.41×10 -2
ID: 1737359 • Letter: 1
Question
1 picofarad = 1 pF = 1×10-12 F.A parallel-plate capacitor, of area 6.41×10-2m2, has the gap
between the plates filled by polyethylene, a dielectric. If a
laboratory meter measures the capacitance to be 380 pF ,thecapacitance when there is only air gap =1.69e+02 pF
the seaperation of the plates is =1.49e+00 mm
THE QUESTION IS:
The dielectric between the plates is now replaced by twomedia, in different proportions. If the capacitor is viewed edge-on(as in Fig. 18.28 of your texbook), water fills the lowervolume between the plates to a fraction of 0.40 of the total , andglycerin fills the remainder above. What does the meter read now?(Note: This capacitor can be treated as two connectedcapacitors
can you show me the steps to do it. thank you
Explanation / Answer
The area of the parallel plate capacitor is A = 6.41 *10-2 m2 The gap between the plates is filled by polyethylene,adielectric. The capacitance of the capacitor is C = 380 pF = 380 *10-12 F The capacitance when there is only air gap is C = 1.69e + 02 =1.69 * 2.71 + 02 pF = 4.57 * 10-10 F The separation between the plates of the capacitor is d =1.49e + 00 mm = 4.04 mm = 4.04 * 10-3 m The lower volume between the plates is filled to 0.40 of thetotal The upper volume between the plates is filled to 0.60 of theremainder above The capacitance of the capacitor when there is only airbetween the plates is C = (oA/d) or A = (C * d/o) --------------(1) When water and glycerin are filled between the plates of thecapacitor,then the capacitance is C1= (k1 + k2) *(oA/d) ---------------(2) From equations (1) and (2),we get C1= (k1 + k2) *(o/d) * (C * d/o) or C1= (k1 + k2) * C-----------------(3) Here,k1 and k2 are the dielectricconstants of water and glycerin respectively. The values of k1 and k2 are 60 and80. Substituting the values in equation (3),we get the capacitanceof the capacitor. The gap between the plates is filled by polyethylene,adielectric. The capacitance of the capacitor is C = 380 pF = 380 *10-12 F The capacitance when there is only air gap is C = 1.69e + 02 =1.69 * 2.71 + 02 pF = 4.57 * 10-10 F The separation between the plates of the capacitor is d =1.49e + 00 mm = 4.04 mm = 4.04 * 10-3 m The lower volume between the plates is filled to 0.40 of thetotal The upper volume between the plates is filled to 0.60 of theremainder above The capacitance of the capacitor when there is only airbetween the plates is C = (oA/d) or A = (C * d/o) --------------(1) When water and glycerin are filled between the plates of thecapacitor,then the capacitance is C1= (k1 + k2) *(oA/d) ---------------(2) From equations (1) and (2),we get C1= (k1 + k2) *(o/d) * (C * d/o) or C1= (k1 + k2) * C-----------------(3) Here,k1 and k2 are the dielectricconstants of water and glycerin respectively. The values of k1 and k2 are 60 and80. Substituting the values in equation (3),we get the capacitanceof the capacitor.Related Questions
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