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Gravitational Potential Energy

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Gravitational Potential Energy Briefly review the concepts of potential energy and work. •Potential Energy = U = stored work in a system •Work = energy put into or taken out of system by forces •Work done by a (constant) force F : v v v v F W = F ⋅∆r =| F || ∆r | cosθ θ ∆r Gravitational Potential Energy Lift a book by hand (Fext) at constant velocity. F = mg final ext Wext = Fext h = mgh h Wgrav = -mgh Fext Define ∆U = +Wext = -Wgrav= mgh initial Note that get to define U=0, mg typically at the ground. U is for potential energy, do not confuse with “internal energy” in Thermo. Gravitational Potential Energy (cont) For conservative forces Mechanical Energy is conserved. EMech = EKin +U Gravity is a conservative force. Coulomb force is also a conservative force. Friction is not a conservative force. If only conservative forces are acting, then ∆EMech=0. ∆EKin + ∆U = 0 Electric Potential Energy Charge in a constant field ∆Uelec = change in U when moving +q from initial to final position. ∆U = U f −Ui = +Wext = −W field FExt=-qE + Final position v v ∆U = −W = −F ⋅∆r fieldv field FField=qE v ∆r → ∆U = −qE ⋅∆r E + Initial position -------------- General case What if the E-field is not constant? v v ∆U = −qE ⋅∆r f v v Integral over the path from initial (i) position to final (f) ∆U = −q∫ E ⋅dr position. i Electric Potential Energy Since Coulomb forces are conservative, it means that the change in potential energy is path independent. f v v ∆U = −q∫ E ⋅dr i Electric Potential Energy Positive charge in a constant field Electric Potential Energy Negative charge in a constant field Observations • If we need to exert a force to “push” or “pull” against the field to move the particle to the new position, then U increases. In other words “we want to move the particle to the new position” and “the field resists”. • If we need to exert a force to “hold” the particle so the field will not move the particle to the new position, then U decreases. In other words, “the field wants to move the particle”, and “we resist”. • Both can be summarized in the following statement: “If the force exerted by the field opposes the motion of the particle, the field does negative work and U increases, otherwise, U decreases” Potential Energy between two point charges +Q r 1 +Q2 Imagine doing work to move the objects from infinitely far apart (initial) to the configuration drawn above (final). Potential Energy between two point charges ∆U = U f −Ui = +Wext = −W field f v v ∆U = U −U = −∫ F ⋅dr * Note that force is not f i field constant over the path! i f v v ∆U = U f −Ui = −∫ qE ⋅dr i Consider Q1 fixed and move Q2 from infinity to r. r +Q1 11 Potential Energy between two point charges +Q r 1 +Q2 f v v E-field generated by Q . ∆U = U −U = − qE ⋅dr 1 f i ∫ Moving Q2 through the field. i r ⎛ ⎞ ⎜ 1 Q1 ⎟ ∆U = U f −Ui = −∫Q2 ⎜ 2 ⎟dx ∞ ⎝ 4πε 0 x ⎠ 1 Q1Q2 ∆U = U f −Ui = 4πε 0 r Potential Energy between two point charges We also need to define the zero point for potential energy. This is arbitrary, but the convention is U=0 when all charged objects are infinitely far apart. 1 Q Q 1 2 U =U(∞)= 0 by our convention. ∆U = U f −Ui = i 4πε 0 r 1 Q1Q2 Potential Energy between two U f = electric charges. 4πε 0 r E. Potential energy vs. distance E. Potential Energy of a charge distribution q1 Potential energy associated to the q2 field produced by charges qi r1 r2 q q ⎛ q q q ⎞ q q 3 U = 0 ⎜ 1 + 2 + 3 + ...⎟ = 0 ∑ i 4πε 0 ⎝ r1 r2 r3 ⎠ 4πε 0 i ri r3 q0 For a continuous charge distribution, replace the sum by an integral E. Potential Energy of a charge distribution q1 To calculate the TOTAL potential q2 energy we have to consider all the fields produced by all the charges r qi 1 r on the other(n-1) charges 2 qj q3 1 qiq j UTotal = ∑ r3 4πε i< j r q0 0 ij For a continuous charge distribution, replace the sum by an integral CPS Question Both cases below are for two kQ Q point charges separated by a U (r) = 1 2 r distance r. Which graph is correct for two negative charges? A) Left Plot, B) Right Plot U(r) U(r) r r Just like a compressed spring stores potential energy. 17 Electric Field +Q r 1 +Q2 Earlier we found that not only using forces, but also electric fields was very useful. v 1 Q Q v 1 Q v v 1 Q 1 2 ˆ 1 1 F12 = 2 r E1 = 2 rˆ F12 = qE1 = Q2 2 rˆ 4πε 0 r 4πε 0 r 4πε 0 r •Electric Field is the force per unit of charge due to the presence of Q1 •Electric field from Q1 is there even if Q2 is not there. All of the above are vectors! Electric Potential +Q r 1 +Q2 We find a similarly useful thing with electric potential energy. 1 Q1Q2 1 Q1 1 Q1 U12 = V1 = U12 = qV1 = Q2 4πε 0 r 4πε 0 r 4πε 0 r •Electric potential is the electric potential energy per unit of charge due to the presence of Q1 •Electric potential from Q1 is there even if Q2 is not there. All of the above are scalars! Electric Potential v Electric Field is a vector associated with 1 Q1 E rˆ a source charge Q1. 1 = 2 +Q1 4πε 0 r Units are [Newtons/Coulomb]. Electric potential is a scalar associated with 1 Q1 a source charge Q . V1 = 1 4πε 0 r Units are [Joules/Coulomb] or [Volts]. V = Voltage = Electric Potential Units are [Volts] = [Joules/Coulomb] U = Electric Potential Energy Units are [Joules] Electric Field Units=[Volts]/[meter]=[Newton]/[Coulomb] Electric Potential of a point charge 1 q V = 4πε 0 r Analogy: Electrical pressure or electrical "height" Positive charges want to get away from higher voltage towards lower voltage. Just like a gas wants to move from high to low pressure. Electric Potential of a charge distribution q1 Potential energy associated to the q2 field produced by the charges qi r 1 r q ⎛ q q q ⎞ q q 2 U = 0 ⎜ 1 + 2 + 3 + ...⎟ = 0 ∑ i 4 r r r 4 r q3 πε 0 ⎝ 1 2 3 ⎠ πε 0 i i Electric potential due to charges q r3 i q0 1 qi “test charge” V = ∑ 4πε 0 i ri For a continuous charge distribution, replace the sum by an integral Electric Potential from Electric Field f r v f v v ∆U = WExt. = −∫ F ⋅ dr = −∫ qE ⋅ dr i i f ∆U v v ∆V = = −∫ E ⋅ dr q i Equipotential surfaces Elevation is a scalar. Contour lines show paths of constant elevation. Gravitational potential VG=gh If I stand at a certain elevation I have a gravitational potential energy [Joules] = mgh=mVG Equipotential surfaces Electric potential (Voltage) is a scalar. Contour lines show paths of constant Voltage (equipotentials). If a charge q is at a certain Voltage, it has an electrical potential energy [Joules] = qV Equipotential surfaces and field lines •Equi-potential surfaces and field lines are always mutually perpendicular •The Field is not necessarily constant on equipotential surfaces •Larger density of equipotentials means larger variations of V, and larger values of |E| E. Field and E. potential What is Electric Field? Electric Field is a vector that is analogous to a steepness vector. Steepness cannot have one number at a given position, it depends which direction you look (vector). E. Field and E. potential f Given the Electric field vector v v as a function of position, we can ∆V = V f −Vi = −∫ E ⋅dr compute Voltage. i r v v V (r) −V (∞) = −∫ E ⋅dr Vi =V(∞)= 0 by our convention. ∞ r v v If we integrate the “steepness” over a path, we V V (r) E dr find the change in elevation. ∆ = = −∫ ⋅ This does not depend on our path taken. ∞ E. Field and E. potential b v v b ∆V = −∫ E⋅dr = −∫ dV a a dV dV dV − dV = E dx + E dy + E dz ⇒ E = − ; E = − ; E = − x y z x dx y dy z dz v The Electric field vector is the rate of change ∂V ∂V ∂V of Voltage in a given direction. E = − xˆ − yˆ − zˆ ∂x ∂y ∂z v r The Electric field vector “potential gradient” E = −∇V http://www.falstad.com/vector2de/ CPS question Two identical charge, +Q and +Q, are fixed in space. The electric potential (V) at the point X midway between the charges is: A) Zero B) Non-Zero V v ∂V E = − xˆ E=0 ∂x V=? +Q +Q Point X +Q +Q x 30 CPS Question Drawn are a set of equipotential lines. Consider the electric field at points A and B. Which of the following statements is true? A)|EA| > |EB| B)|E | < |E | Point B A B C)|E | = |E | 0V 10V 20V 30V A B Point A D)Not enough information given. E) None of the above CPS Question Two charges, +Q and -Q, are fixed in space. The electric field at the point X midway between the charges is: A) Zero B) Non-Zero V v ∂V E = − xˆ V=0 ∂x E=? +Q -Q +Q -Q Point X x 32 CPS Question The magnitude of the electric field at point P is: +Q +Q A) Zero B) Non-Zero Point P The magnitude of the voltage at point P is: A) Zero B) Non-Zero -Q -Q 33 E.
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