PHY 2049: Physics II We begin with a live clicker today. Wileyplus homework should be fully operational. Tea and Cookies: We meet on Tuesdays at 4:00PM for tea and cookies in room 2165.
quiz The electric field at a distance of 1 m from an isolated point particle with a charge of 2×10−9 C is: A. 1.8N/C B. 180N/C C. 18N/C D. 1800N/C E. none of these c
Quiz 2 An isolated charged point particle produces an electric field with magnitude E at a point 2m away from the charge. A point at which the field magnitude is 4E is: A. 1m away from the particle B. 0.5m away from the particle C. 2m away from the particle D. 4m away from the particle E. 8m away from the particle
PHY 2049: Physics II Last week Coulomb’s law, Electric Field and Gauss’ theorem Today Electric Potential Energy and Electric Potentials Numerous cases
Potential Energy and Potential Force => work => change in K=> change in Potential energy ΔU = Uf – Ui = -W = - ΔK Work done is path independent. K+U = constant. U = k q1 q2/r : interaction energy of two charges. Sign matters
PHY 2049: Physics II Electric Potential Also V = U/q = -W/q Units of Joules/coulomb = volt 1 eV = e x 1V = 1.6 x 10-19 J Also V = kq/r Vf –Vi = -∫E.ds In case of multiple charges, add as a number
Neg., lower ? Pos., higher Neg., higher PHY 2049: Physics II Conservation of energy tell us that Positive charge moves downhill (its kinetic energy increases) and negative charge moves uphill. Can you tell the sign of the charge by looking at its behavior over a surface of potentials. Is the speed at the end bigger or smaller than in the beginning.
PHY 2049: Physics II Vf-Vi = -∫ k q/r2 dr Choose Vi = V (∞)=0 V(r) = kq/r
Uniformly charged disk V = kpcosθ/r2 E = -∂V/∂s = - Uniformly charged disk V = ??
A B C (24 - 6)
dq O A (24 - 8)
PHY 2049: Physics II
. P r dq (24 - 7)
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S A B V q S V (24 - 11)
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(24 - 13) A B V V+dV
(24 - 14) A B V V+dV
(24 - 15) x y O q1 q2 q3 r12 r23 r13
O x y q1 Step 1 (24 - 16) x O r12 q1 q2 Step 2 Step 3 x y O r12
(24 - 17) path A B conductor
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(24 - 20) R