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9. Equipotential Surfaces Equipotential Surfaces are always perpendicular to Electric Field Lines The surface of a conductor is ALWAYS an Equipotential surface 10. Capacitance C = Q/V Capacitance has units of “farads” 1 farad = 1 Coulomb/Volt Q A C= =ε 0 Vab d Recall result from infinite flat plate Integral is easy because field is uniform Capacitance between two parallel plates” a r b r Va − Vb = ∫ E ⋅ dl = Ed 1 Qd σ Q = E= ε0 ε0 A V = Ed = ab ε0 A +Q -Q To add more charge, I must do Work Energy Storage in a Capacitor Q V= C To add a small amount of charge dq I must do work dW - q dW = Vdq = dq C Q final Q final Qinitial Q final ∫ dW = ∫ Vdq = ∫ Qinitial Stored Energy in a Charged Capacitor 1 2 ⎞ 1 ⎛1 1 q qdq = ⎜ Q 2final = Qinitial dq = ⎟ ∫ 2 C⎝2 C Qinitial C ⎠ Q final To add a finite charge Q’ I must integrate - W= Qinitial If I start from NO net charge I get - 1 1 Q2 1 W= = QV = CV 2 2 C 2 2 Capacitors in Parallel Q2 = C2V Q1 = C1V Q = Q1 + Q2 = (C1 + C2 )V Q = C1 + C2 V Cequiv = C1 + C2 Cequiv = C1 + C2 + C3 + .... Capacitors in Series Vac = V1 = Q C1 Vcb = V2 = ⎛ 1 1 ⎞ Vac = V = V1 + V2 = Q⎜⎜ + ⎟⎟ ⎝ C1 C2 ⎠ V 1 1 = + Q C1 C2 1 Cequiv 1 Cequiv = 1 1 + C1 C2 = 1 1 1 + + + .... C1 C2 C3 Q C2 11. Resistance and Ohms Law Definition: A Resistor is a device that follows Ohms Law: V=IR If there is a voltage “across” a resistor then There MUST be a current “through” it. If there is a current “through” a resistor then There must be a voltage across it. Step 2- Label ALL Currents – Use Kirchhoff’s Junction Rule to simplify The directions of currents you chose is NOT IMPORTANT If you guess wrong – Mr. Kirchhoff will give you a minus sign for the current. Step 5 – Go around the loops to get the equationsNOTE – If we go from + to – potential change is NEGATIVE If we go from - to + potential change is POSITIVE + + + + + 13V − 13V − I 1 (1Ω ) − (I 1 − I 3 )(1Ω ) = 0 − I 2 (1Ω ) − (I 2 + I 3 )(2Ω ) + 13V = 0 − I 1 (1Ω ) − I 3 (1Ω ) + I 2 (1Ω ) = 0 Loop 1 Loop 2 Loop 3 Step 6- Solve the Simultaneous Equations 13V − I 1 (1Ω ) − (I 1 − I 3 )(1Ω ) = 0 Loop 1 − I 2 (1Ω ) − (I 2 + I 3 )(2Ω ) + 13V = 0 Loop 2 − I 1 (1Ω ) − I 3 (1Ω ) + I 2 (1Ω ) = 0 Loop 3 Solve Equation 3 for I2: I 2 = I1 + I 3 Plug into Equations 1 and 2 (this gives 2 equations and 2 unknowns): 13V = I 1 (2Ω ) − I 3 (1Ω ) Note Y&F have misprint here p991 13V = I 1 (3Ω ) + I 3 (5Ω ) Multiply top equation by 5 and add two equations: 65V = I 1 (10 Ω ) − I 3 (5Ω ) 13V = I 1 (3Ω ) + I 3 (5Ω ) 78V = I 1 (13Ω ) I1 = 6 A I 3 = −1 A Negative sign here means we guessed wrong about direction of I3 I2 = 5 A