mat-sln-asn-hwk10-spr02

# mat-sln-asn-hwk10-spr02 - Solution for Homework 10...

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Unformatted text preview: Solution for Homework 10 Computing with Capacitance Solution to Homework Problem 10.1(Energy of Charged Sphere as Radius is Changed) Problem: How does the energy of a uniformly charged spherical shell change as the radius of the sphere is increased without changing the total charge? Select One of the Following: (a) Increases. (b-Answer) Decreases. (c) Stays the same. (d) Increases then Decreases. (e) The sign of the change in energy cannot be determined. Solution An external agent would have to do positive work on the system to make the sphere smaller since the agent would have to exert a force to move the charges closer together. Therefore the energy of the system decreases as the system gets larger. Total Points for Problem: 3 Points Solution to Homework Problem 10.2(Potential Difference Non-Uniform Charge) Problem: A NON-UNIFORM spherical volume charge with volume charge density ρ ( r ) = γr 4 occupies the spherical region < r < a where γ is a positive constant. Compute the potential difference Δ V a between the outer edge of the charge and the center of the charge. Be careful of the sign. Select One of the Following: (a) γr 6 18 ε (b) 5 γa 7 13 rε (c) 3 γr 6 28 ε (d) 7 γa 6 15 ε (e-Answer) γa 6 42 ε Solution (a) Compute Charge Enclosed by a Spherical Gaussian Surface: A Gaussian surface with a radius r < a encloses a total charge Q enc = integraldisplay r 4 πr 2 ρdr = integraldisplay r 4 πr 2 γr 4 dr = 4 πγ integraldisplay r r 6 dr Perform the integral Q enc = 4 πγ r 7 7 vextendsingle vextendsingle vextendsingle vextendsingle r = 4 πγ r 7 7 1 (b) Compute the Field Using Gauss’ Law: For a spherically symmetric system, Gauss’ law can be reduced to E = Q enc 4 πε r 2 ˆ r = 4 πγ r 7 7 4 πε r 2 ˆ r = γr 5 7 ε ˆ r (c) Compute the Potential Difference: Since the field depends on only one variable r , the potential can be found by an indefinite integral, V ( r ) =- integraldisplay Edr =- integraldisplay γr 5 7 ε dr =- γr 6 42 ε + C where C is the constant of integration. The potential difference is then Δ V a = V (0)- V ( a ) = C- parenleftbigg- γa 6 42 ε + C parenrightbigg = γa 6 42 ε Total Points for Problem: 3 Points Solution to Homework Problem 10.3(Energy Stored in Small Lab Capacitor) Problem: Last semester, a couple of students were winding new Gauss Guns for me (plumbing parts and 14 guage), and we blew up 4 of our high voltage power supplies. To test the power supplies, I began charging one of small blue capacitors we used in the RC lab to 50V . I was quite surprised that a very significant spark was generated when I discharged the capacitor. The capacitor has a capacitance of 1000 μ F . How much energy did I store in the capacitor?...
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mat-sln-asn-hwk10-spr02 - Solution for Homework 10...

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