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i safari File Edit View History Bookmarks Window Help 83% , Tue 8:33 PM a ezto.mheducation.com My.UTEP Lecture 9 - PDF-C https://blackboard https://blackboardl Mail - abeard2@mi. UTEP Prof. Dirk Exa.. Homework Chapter. Physics question hot PHYS1404 Genenal Physics Il: Fall 2017 Alexandra Beard 34 PM YSICS Homework Chapter 17 instructions I help hot Question 11 (of 12) Save & Exit Submit 4 PM 10.00 points 11 PM 1 out of 8 attempts To make a parallel plate capacitor, you have available two flat plates of aluminum Assistance (are.-1902hash et ofpaper (thickness Check My Work View Hint View Question Show Me Guided Solution Practice This Question Print Question Help Report a Problem 0.10 mm, K-35), asheet ofglass (thickness 2.0 mm,K = 7.0), and a slab of paraffin (thickness = 10.0 mm, K-2.0). (a) What is the largest capacitance possible using one of these dielectrics? hot 51 PM nF (b) What is the smallest? pF ot 01 PM ot 51 PM References eBook & Resources 26

Explanation / Answer

here,

area = 0.019 m^2

capacitanace , C = area * k * e0 /d

as ratio of K/d (r = K/d) increases the capacitance increases

for sheet of paper ( d = 0.0001 m , k = 3.5)

r1 = 3.5 /0.0001 = 3.5 * 10^4 m^-1

for sheet of glass ( d = 0.002 m , k = 7)

r2 = 7 /0.002 = 3.5 * 10^3 /m

for slab of paraffin ( d = 0.01 m , k = 2)

r3 = 2 /0.01 = 200 m^-1

a)

for largest capcitance ,

we use sheet of paper ,

C = area * e0 * r1

C = 0.019 * 8.85 * 10^-12 * 3.5 * 10^4 F

C = 5.89 * 10^-9 F = 5.89 nF

b)

for smallest capacitance

we use Paraffin slab

C = area * e0 * r3

C = 0.019 * 8.85 * 10^-12 * 200 F

C = 33.6 pF

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