LECT5_161165

# LECT5_161165 - Potential Energy of a system of charges q d...

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Potential Energy of a system of charges Potential Energy PE (scalar): Δ PE = – Work done by the Electric field + + + + + + + _ _ _ _ _ _ _ +q +q d F = qE Δ PE= –W= – Fd= – qEd (units = J) Work done by the E-field (to move the +q closer to the negative plate) REDUCES the P.E. of the system If a positive charge is moved AGAINST an E-field (which points from + to -), the charge-field system gains Pot. Energy. If a negative charge is moved against an E-field, the system loses potential energy

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Electric Potential Difference, Δ V Δ V = V B – V A = Δ PE / q Units: Joule/Coulomb = VOLT Scalar quantity Relation between Δ V and E: Δ V = Ed E has units of V/m = N/C (V / m = J / Cm = Nm / Cm = N / C) + + + + + + + _ _ _ _ _ _ _ +q Point A +q Point B d
Potential vs. Potential Energy POTENTIAL: Property of space due to charges; depends only on location Positive charges will accelerate towards regions of low potential. POTENTIAL ENERGY: due to the interaction between the charge and the electric field + + + + + + + _ _ _ _ _ _ _ +q +q PE 1 PE 2 Δ PE + + + + + + + _ _ _ _ _ _ _ V 1 V 2 Δ V

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A parallel plate capacitor has a constant electric field of 500 N/C; the plates are separated by a distance of 2 cm. Find the potential difference between the two plates. Example of Potential Difference + + + + + + + _ _ _ _ _ _ _ Δ V Remember: potential difference Δ V does not depend on the presence of any test charge in the E-field! d E-field is uniform, so we can use Δ V = Ed = (500V/m)(0.02m) = 10V
Now that we’ve found the potential difference Δ V, let’s take a molecular ion, CO 2 + (mass = 7.3 × 10 –26 kg), and release it from rest at the anode (positive plate). What’s the ion’s final velocity when it reaches the cathode (negative plate)? Example of Potential Difference + + + + + + + _ _ _ _ _ _ _ Δ PE Solution: Use conservation of energy: Δ PE = Δ KE Δ PE = Δ V q Δ KE = 1/2 m v final 2 – 1/2 m v init 2 Δ V q = 1/2 m v final 2 v final 2 = 2 Δ Vq/m = (2)(10V)(1.6 × 10 –19 C)/7.3 × 10 –26 kg v final = 6.6 × 10 3 m/s

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Thunderstorms: From ground to cloud base: Δ V = 10 8 V, E ~ 10 4-5 V/m Lightning: E = 3 × 10 6 V/m is electric field strength at which air becomes ionized enough to act as a conductor. Fair weather: E ~ 10 2 V/m
Example for Potential Difference: BATTERIES: the potential difference across terminals is kept at a constant value, e.g., 9 volts

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Recall that for a single point charge. Potential
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LECT5_161165 - Potential Energy of a system of charges q d...

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