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PHYS219 Fall Semester 2014 Lecture 4: Gauss Law & Potential Energy Dimitrios Giannios Purdue University Help Center Info (see syllabus link) Remember: Homework 1 is due this Friday, 10:30am before the recitation! Electric Flux and Gauss’ Law • Gauss’ Law can be used to find the electric field of a complex charge distribution • Easier than treating it as a collection of point charge and using superposition • To use Gauss’ Law, a quantity called electric flux is needed • The electric flux is equal the product of the electric field that passes through a particular surface and the area of the surface • The electric flux is denoted by E The idea of an electric flux How do you account (in a very general mathematical way) for the E fields that fill space? How many E-field lines penetrate a small surface with area dA? Definition of the electric flux, E||=|E|sin E =|E|cos E Must be able to visualize the components of E n̂ DEFINE E =dA x E = dA E cos θ [units]: Nm2/C Area dA Special cases: Gauss’ Law Think about closed surfaces: two possibilities E ,total≈ q dA dA What can be shown: For ANY closed surface ΦE,total=closed surfacedA E cos θ = 4 π k qenclosed dA when no net charge lies inside the surface define k 1 4 π ε0 ε0 =8.85 x 10-12 C2 / m2 N qenclosed Gauss ' s Law: E ,total ε0 Applications of Gauss Law Calculate the electric field for charge configurations with some symmetry: Gauss’ Law: Point Charge • Choose a Gaussian surface • Want a surface that will make the calculation as easy as possible • Choose a surface that matches the symmetry of the problem • For a point charge, the field lines have a spherical symmetry Gauss’ Law: Point Charge (more) • Since the field is perpendicular to the area, the flux is • • • • the product of the field and the area: E = E Asphere Asphere is the area of the Gaussian sphere With a radius of r, Asphere = 4 r2 Therefore, E = 4 r2 E From Gauss’ Law, FE = 4p r E = 2 q eo and E= q 4pe o r 2 • This agrees with the result from Coulomb’s Law Applications of Gauss law will follow in next lectures Part II: Electric Potential Energy • Electric forces can do work on a charged object • Work is related to changes in electric potential energy • Analogous in many ways to gravitational potential energy • Conservation of energy will be revisited • The ideas of forces, work, and energy will be extended to electric forces and systems Electric Potential Energy • A point charge in an electric field experiences a force: • Assume the charge moves a distance Δx • The work done by the electric force on the charge is W = FE Δx • The electric force is conservative, so the work done is independent of the path • If the electric force does an amount of work W on a charged particle, there is an accompanying change in electric potential energy Electric Potential Energy, cont. • The change in electric potential energy is Δ(PE)E = - W = - FE Δx = -q E Δ x • The change in potential energy depends on the endpoints of the motion, but not on the path taken Δ(PEE) + W=0 Notation: PE=Potential Energy (J) KE=Kinetic Energy (J) W=work (J) Gravitational Potential Energy (PEG) Electrostatic Potential Energy (PEE) end end Δr Δh begin + m q1 + Conventions: near earth’s surface, PEG=0 for orbiting masses, PEG=0 at infinity earth + q2 begin “fixed” PEE=0 (by definition) when separation is at infinity “fixed” Electrostatic potential energy is the energy stored in a system of charges. It depends on the relative positions (not the motion) of the charges in the system. Gravity and Electrostatics What is the same, what’s different between What’s the Same: PE can be converted to KE and vice versa. PE can be converted to W and vice versa. What’s Different: • Gravity is attractive: Gravitational Potential Energy is negative (when set to 0 at infinite distance • Electrostatic Potential Energy both positive and negative (depending on the sign of charges) Bottom Line: The situation for electrostatics is similar to gravity, except you can’t see the hills and valleys. Need an imagination! Review: Gravitational Analogy (Potential Energy is Stored Energy) PEG increases Startin g position m Reference Level Hill Δ(PEG)+W=0 Work done on the ball It’s basically Conservation of Energy Hole PEG decreases Δ(PEG)+W=0 Work done by the ball CASE I: Change in Electrostatic Potential Energy (case of like charges) end PEE Must push “+” charge with external force Notes: •The change in PEE is a positive Q number •Charges are stationary at beginning and end (KE=0) •Same diagram if both charges are negative begin + + PEE increases, W is negative r + q An amount of work W is required Δ(PEE) + W=0 The work done by the electric force is negative Electrostatic Potential Energy (case of unlike charges) CASE II: Change in PEE Note that the change in PEE is a positive number Q Must push “+” charge with external force - + PE increases Charges are stationary at beginning and end (KE=0) begin PEE increases, W is negative r end + q An amount of work W is required Δ(PEE) + W=0 13e electric The work is done by th force is negative Electrostatic Potential Energy (more) Now return the charge to original position. What is the change in Electrostatic Potential Energy? PEE Charges are stationary at beginning and end (KE=0) Q Must hold back “+” charge with external force r + - + PE decreases begin + q Work will be recovered Δ(PEE) + W=0 end PEE decreases, W is positive We say work is done by the electric fields of the charges Making the Discussion Quantitative (point charges ONLY) ri rf Move charge q2 from ri to rf What is the PEE stored in the system? ri infinity PEE,i = 0 Convention: PEE is zero at infinite distance 0 PEE D(PEE ) = PEE,f - PEE,i separation kq1q2 kq1q2 = rf ri PEE,f kq1q2 = rf 0 NOTE It is important to use the signs associated with each charge!!! When the signs of the two charges are opposite, the potential energy of the pair is negative. When the signs of the two charges are the same, the potential energy of the pair is positive. kq1q2 PEE = r An example -2 C kq1q2 PEE = r 3 C - + PEE (r Electrostatic Potential Energy PEe 2.5 m q1 - 2.5m) = -2.16 x1010 J What does the minus sign mean?? q2 + r Extracting PE from a system wrt infinite separation indicates a preference toward confinement, like digging a hole on the beach in gravitational situation. PEelec and Superposition • The results for two point charges can be extended by using the superposition principle • If there is a collection of point charges, the total potential energy is the sum of the potential energies kq1q2 kq2q3 kq1q3 of each pair of charges PEE = + + r12 r23 r13 question: how many potential terms do you have for 4 charges? 1+2+3 = 6 terms (n-1) x n/2 for n charges