# E2HWCH24 - Chapter 24 Capacitance 5 •[SSM A...

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Unformatted text preview: Chapter 24 Capacitance 5 • [SSM] A parallel-plate capacitor is connected to a battery. The space between the two plates is empty. If the separation between the capacitor plates is tripled while the capacitor remains connected to the battery, what is the ratio of the final stored energy to the initial stored energy? Determine the Concept The energy stored in a capacitor is given by QV U 2 1 = and the capacitance of a parallel-plate capacitor by . d A C ∈ = We can combine these relationships, using the definition of capacitance and the condition that the potential difference across the capacitor is constant, to express U as a function of d. Express the energy stored in the capacitor: QV U 2 1 = (1) Use the definition of capacitance to express the charge of the capacitor: CV Q = Express the capacitance of a parallel-plate capacitor in terms of the separation d of its plates: d A C ∈ = where A is the area of one plate. Substituting for Q and C in equation (1) yields: d AV U 2 2 ∈ = Because d U 1 ∝ , tripling the separation of the plates will reduce the energy stored in the capacitor to one-third its previous value. Hence the ratio of the final stored energy to the initial stored energy is 3 / 1 . 16 • An isolated conducting sphere that has a 10.0 cm radius has an electric potential of 2.00 kV (the potential far from the sphere is zero). ( a ) How much charge is on the sphere? ( b ) What is the self-capacitance of the sphere? ( c ) By how much does the self-capacitance change if the sphere’s electric potential is increased to 6.00 kV? Picture the Problem The charge on the spherical conductor is related to its radius and potential according to V = kQ / r and we can use the definition of capacitance to find the self-capacitance of the sphere. ( a ) Relate the potential V of the spherical conductor to the charge r kQ V = ⇒ k rV Q = on it and to its radius: Substitute numerical values and evaluate Q : ( 29 ( 29 nC 3 . 22 nC 252 . 22 C m N 10 8.988 kV 2.00 cm . 1 2 2 9 = = ⋅ × = Q ( b ) Use the definition of capacitance to relate the self-capacitance of the sphere to its charge and potential: pF 11.1 kV 2.00 nC 22.252 = = = V Q C ( c ) It doesn’t. The self-capacitance of a sphere is a function of its radius. 23 •• [SSM] An air-gap parallel-plate capacitor that has a plate area of 2.00 m 2 and a separation of 1.00 mm is charged to 100 V. ( a ) What is the electric field between the plates? ( b ) What is the electric energy density between the plates? ( c ) Find the total energy by multiplying your answer from Part ( b ) by the volume between the plates. ( d ) Determine the capacitance of this arrangement....
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E2HWCH24 - Chapter 24 Capacitance 5 •[SSM A...

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