hw4.6 - b A a c A b A a A C C C C C B A 1 1 1 1 1 1 1 c b a...

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HW4.6 Given: 7 . 4 Find: ? C 2 4 . 2 m A ? U cm c a 8 . 1 ? C cm b 8 . 1 ? C v 7 V 5 . 2 A parallel plate capacitor is shown in the diagram in cross section. A slab of dielectric ( = 4.7) is positioned midway between the plates. The area of the plates and dielectric slab is A = 2.4 m 2 . The spacings are a = c = 1.8 cm and b = 1.8 cm. (a) Find the capacitance. (b) A battery with E = 7 V is now connected across the capacitor. How much energy is stored in the capacitor when fully charged? (c) The dielectric slab is removed and replaced by a slab of conducting material with identical dimensions. The spacings a , b , and c are unchanged. Calculate the capacitance now. (d) If slabs of dielectric with ' = 2.5 are next inserted, filling the empty space between the top plate and conducting slab and the empty space between the lower plate and the conducting slab, what is the capacitance of the resulting configuration? Solutions a.) To find C we can use the following formula: d A C 0 0 0 C C a A C A 0 ; b A C B 0 ; c A C C 0 They are in series A c b a A c A
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Unformatted text preview: b A a c A b A a A C C C C C B A 1 1 1 1 1 1 1 c b a A C b.) To find U we can use the following formula: 2 2 2 1 2 1 V c b a A CV U c b a AV U 2 2 c.) To find C we can use the following formula: a A C A ; c A C C A c a A c A a c A a A C C C C A 1 1 1 1 1 b a A C d.) To find C we can use the following formula: a A C A ; c A C C A c a A c A a c A a A C C C C A 1 1 1 1 1 b a A C...
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hw4.6 - b A a c A b A a A C C C C C B A 1 1 1 1 1 1 1 c b a...

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