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Electric Potential • Hillside analogy • Constant electric fields • Work and PE moving in constant field • PE and Electric Potential in constant field • PE and Electric Potential examples Topographical Analogy (1) • Consider a hillside – top and side view • 2 quantities – Elevation – Slope • 2 are related – 𝑠𝑙𝑜𝑝𝑒 = ∆𝑒𝑙𝑒𝑣 ∆𝑥 – ∆𝑒𝑙𝑒𝑣 = 𝑠𝑙𝑜𝑝𝑒 ∙ ∆𝑥 Topographical Analogy (2) • 2 are related – 𝑠𝑙𝑜𝑝𝑒 = ∆𝑒𝑙𝑒𝑣 ∆𝑥 – ∆𝑒𝑙𝑒𝑣 = 𝑠𝑙𝑜𝑝𝑒 ∙ ∆𝑥 • Observe: – When elevation changes rapidly, slope is steep! – When elevation changes gradually, slope is gradual! Topographical Analogy (3) • Another point: – Slope can be different in different directions (like a vector) – 𝑠𝑙𝑜𝑝𝑒𝑥 = – 𝑠𝑙𝑜𝑝𝑒𝑦 = ∆𝑒𝑙𝑒𝑣 ∆𝑥 ∆𝑒𝑙𝑒𝑣 ∆𝑦 – Elevation is just a number (like a scalar) Example - Mt Gretna Topo Map Constant elevation lines When lines are close together – its steep! Topographical/Electrical comparison • Elevation -> Electric Potential – Scalar quantity V – No direction, just number – Units Joules/Coulomb (volts) • Slope -> Electric Field – – – – Vector quantity E Components in x,y,z direction Units Newtons/Coulomb or Volts/m • 𝐸𝑥 = ∆𝑉 ∆𝑥 𝐸𝑦 = ∆𝑉 ∆𝑦 Electric Potential and Field Map • Highest potential near (+) charge, lowest potential near (-) charge • Equipotential lines (green) analogous to constant elevation lines. • Electric field lines (red) perpendicular, point straight down “hill” Electric Field / Potential example 5. How strong is the electric field between two parallel plates 6.0 mm apart if the potential difference between them is 110 V? 𝑉 110 𝑉 𝑉 𝐸 = 𝑑 = 0.006 𝑚 = 18333 𝑚 = 18333 𝑁𝑚 𝐶 𝑚 𝑁 = 18333 𝐶 6. An electric field of 640 V/m is desired between two parallel plates 11.0 mm apart. How large a voltage should be applied? 7. The electric field between 2 parallel plates connected to a 45-V battery is 1500 V/m. How far apart are the plates? Electric Field vs Force (+/-sign) • Given Electric Field E, how do you find force F? 𝐹 = 𝑞𝐸 Sign of charge flips direction for (+) (-) charges (Electrons feel force “up” the hill) - + Electric Potential vs Potential Energy (+/- sign) • Given ΔV, how do you find Work (ΔPE)? ∆𝑃𝐸 = 𝑞∆𝑉 Sign of charge flips ΔPE for (+) (-) charges (Electrons want to fall “up” the hill) +++++++ +++++++ + PE ∆V ------- PE ∆V ------- Work crossing Electric Potential - Examples • Positive test charge • Negative test charge • Change in Potential ∆𝑉 = −𝐸 ∙ ∆𝑥 • Change in Potential Energy ∆𝑃𝐸 = 𝑞∆𝑉 Electrostatic and Gravitational analogy • Comparison of height and gravitational PE, with electric potential and electrostatic PE Change in Potential ∆𝑉 = −𝐸 ∙ ∆𝑥 • At least rocks don’t fall up! Change in Potential Energy ∆𝑃𝐸 = 𝑞∆𝑉 Example – work moving proton • What is change in PE of a proton moving from +125 V to -55 V? What is change in PE? How much work is done? ∆𝑃𝐸 = 𝑞∆𝑉 𝐽 𝐶 = 1.6 ∙ 10−19 𝐶 (−180 ) = -2.88 ∙ 10−17 Velocity crossing Electric Potential • ΔPE for electron ∆𝑃𝐸 = 𝑞∆𝑉 = −1.6 ∙ 10−19 𝐶 (+5000 𝐽 𝑉) = −8 ∙ 10−16 𝐽 • Change in KE 1 2 𝑚𝑣 2 = −∆𝑃𝐸 = 8 ∙ 10−16 𝐽 𝑣= 2∙8∙10−16 𝐽 9.1∙10−31 𝑘𝑔 = 4.2 ∙ 107 𝑚/𝑠 Change in Potential Energy ∆𝑃𝐸 = 𝑞∆𝑉 Example problems - 1 1. How much work does the electric field do in moving a -7.5 µC charge from ground to a point whose potential is +65 V higher? ∆𝑃𝐸 = 𝑞𝑉 = −7.5𝜇𝐶 (+65 𝐽 𝐶) = −0.00049 𝐽 𝑊 = ∆𝐾𝐸 = −∆𝑃𝐸 = 0.00049 𝐽 2. How much work does the electric field do in moving a proton from a point with a potential of +113 V to a point where it is -45 V? Express your answer both in joules and electron volts. ∆𝑃𝐸 = 𝑞𝑉 = 1.6 ∙ 10−19 𝐶 (−158 𝐽 𝐶) = −2.53 ∙ 10−17 𝐽 𝑊 = ∆𝐾𝐸 = −∆𝑃𝐸 = 2.53 ∙ 10−17 𝐽 2.53∙10−17 𝐽 1.6∙10−19 𝐽 𝑒𝑣 = 158 𝑒𝑣 Example problems - 2 4. An electron acquires 3.51 ✕ 10-16 J of kinetic energy when it is accelerated by an electric field from plate A to plate B. What is the potential difference between the plates, and which plate is at the higher potential? ∆𝑃𝐸 = −∆𝐾𝐸 = −3.51 ∙ 10−16 𝐽 −3.51 ∙ 10−16 𝐽 = ∆𝑃𝐸 = 𝑞𝑉 = −1.6 ∙ 10−19 𝐶 (𝑉) 𝑉= 3.51∙10−16 𝐽 1.6∙10−19 𝐶 = 2193 𝑉 Toward + electrode Electron-volt (ev) • Definition 1 ev is energy electron picks up going across potential difference of 1 volt. ∆𝑃𝐸 = 𝑞𝑉 = −1.6 ∙ 10−19 𝐶 (+1 𝐽 𝐶) = −1.6 ∙ 10−19 𝐽 ∆𝐾𝐸 = −∆𝑃𝐸 = 1.6 ∙ 10−19 𝐽 1 𝑒𝑣 = 1.6 ∙ 10−19 𝐽 Example problems - 3 12. What is the speed of a proton whose kinetic energy is 2.8 keV? KE = 2.8 𝑘𝑒𝑉 = 2800 𝑒𝑣 ∙ 1.6 ∙ 10−19 𝐽 𝑒𝑣 = 4.48 ∙ 10−16 𝐽 1 1 4.48 ∙ 10−16 𝐽 = 𝐾𝐸 = 2 𝑚𝑣 2 = 2 (1.67 ∙ 10−27 𝑘𝑔)𝑣 2 𝑣 = 7.32 ∙ 105 Example problems - 4 12. The work done by an external force to move a -7.50 µC charge from point a to point b is 25.0 ✕ 10-4 J. If the charge was started from rest and had 4.85 ✕ 10-4 J of kinetic energy when it reached point b, what is the magnitude of the potential difference between a and b? 𝑊𝑒𝑥𝑡 = ∆𝐾𝐸 + ∆𝑃𝐸 25 ∙ 10−4 𝐽 = 4.85 ∙ 10−4 𝐽 + ∆𝑃𝐸 ∆𝑃𝐸 = 20.15 ∙ 10−4 𝐽 • Must increase PE by 20.15 ∙ 10−4 𝐽 20.15 ∙ 10−4 𝐽 = ∆𝑃𝐸 = 𝑞𝑉 20.15 ∙ 10−4 𝐽 = −7.5 𝜇𝐶 𝑉 𝑉= 20.15∙10−4 𝐽 −7.5∙10−6 𝐶 = −269 𝑉 • Must move toward negative electrode