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19.1 Electrical Potential Energy Special Case 2 r Electric field created by point charge q1: (diagram drawn assuming q1>0) (1) Direction: pointing directly away from q1 at any point. (2) Magnitude given by Coulomb’s Law E1 ( r ) = k “(r)” means “function of distance r” q1 r 2 (1) *** EPE of test charge q2 in the electric field of source charge q1: q1q2 EPE 21 ( r ) = k r (2) This formula works even if one or both charges are negative 1 ***Derivation of (2) from (1) requires integral calculus 19.1 Electrical Potential Energy Example: An electron (q=−e) is sent, from very far away, towards a charged Q = −0.25 nC. It gets to a distance of 1.10 mm from the center of the sphere before turning around. Q = −0.25 nC What is the initial speed of the electron? electron rf = 1.1 × 10 −3 m ri = ∞ vf = 0 vi = ? e=1.6x10-19C, me=9.11x10-31 kg, k=8.99x109 Nm2/C2 19.1 Electrical Potential Energy Example: An electron (q=−e) is sent, from very far away, towards a charged Q = −0.25 nC. It gets to a distance of 1.10 mm from the center of the sphere before turning around. Q = −0.25 nC What is the initial speed of the electron? E f = Ei → KE f + EPE f = KE i + EPE i 1 2 1 2 mv 2f + m ( 0) 2 + kQqe 1 2 kQqe = 2 mvi + rf ri kQ ( −e) 1 2 kQ ( −e) = 2 mvi + rf ∞ rf = 1.1 × 10 −3 m ri = ∞ vf = 0 vi = ? e=1.6x10-19C, me=9.11x10-31 kg, k=8.99x109 Nm2/C2 → kQ ( − e) 1 2 = 2 mvi rf 2 9 Nm −10 −19 ( )( 2 8.99 × 10 − 2 . 5 × 10 C − 1 . 6 × 10 C) 2 C 2kQ ( −e) 2 ⇒ vi = = (9.11 × 10 −31 kg)(1.1 × 10 −3 m ) mrf m2 = 7.18 × 10 s2 14 ⇒ vi = 2.68 × 10 7 m s electron 19.1 Electrical Potential Energy EPE of qi, qj Total Potential energy of a system of point charges We start with three point charges: q1, q2, q3 U 12 = kq1q2 r12 q1 U ij = q2 r12 U = U 12 + U 13 + U 23 U 13 = kq1q3 r13 r13 r23 U 23 = kqi q j rij kq2 q3 r23 q3 The overall potential energy is the amount of work required of an external agent to assemble this set of charges—this can be done in different ways, for example: (a) Bring in q1 from ∞ (no cost), W1=0 (b) Bring in q2 from ∞: W2=U12 (c) Bring in q3 from ∞: W3=U13+U23 Total W1+ W2+ W2 U=U12+U13+U23 Can do this in any order and obtain the SAME result Extending to N point charges N i −1 U = ∑∑ i =1 j =1 kqi q j rij = 1 2 ∑ j ≠i kqi q j rij There are N(N-1)/2 distinct pairs 4 19.2 The Electric Potential Difference DEFINITION OF ELECTRIC POTENTIAL The electric potential at a given point is the electric potential energy of a small test charge divided by the charge itself: EPE SI Unit of Electric Potential: V= qo joule/coulomb = volt (V) (∆V )AB = VB − VA = Electrical Potenetial Difference from A to B =difference, per unit charge, of the EPE of a test charge WOULD HAVE from location A to location B EPE B EPE B − WAB − = qo qo qo A rA r EPE A = kq1q2 kq , VA = 1 rA rA kq1 V (r) = r rB B EPE B = kq1q2 kq , VB = 1 rB rB 5 19.3 The Electric Potential Difference Created by Point Charges Example The Potential of a Point Charge Using a zero reference potential at infinity***, determine the amount by which a point charge of 4.0x10-8C alters the electric potential at a spot 1.2m away when the charge is (a) positive and (b) negative. 6 19.3 The Electric Potential Difference Created by Point Charges Example The Potential of a Point Charge Using a zero reference potential at infinity***, determine the amount by which a point charge of 4.0x10-8C alters the electric potential at a spot 1.2m away when the charge is (a) positive and (b) negative. (a) kq = r 8.99 ×109 N ⋅ m 2 C 2 + 4.0 ×10 −8 C 1.2 m = +300 V V= ( (b) )( ) kq = V = r (8.99 × 109 N ⋅ m 2 C 2 )(− 4.0 × 10 −8 C) 1.2 m = −300 V *** In Ch. 6 you learned that the choice of the reference zero for gravitational potential energy is arbitrary. This is also true for EPE and hence for the electric potential V. The choice of zero at infinity is only valid for a collection of point charges, and not for parallel plate capacitors or infinite wires 7 19.3 The Electric Potential Difference Created by Point Charges a Example Where is the Potential equal to Zero? x 0 Two point charges are fixed in place. The positive charge is +2q and the negative charge is –q. On the line that passes through the charges, how many places are there at which the total potential is zero? Where are they? 8 19.3 The Electric Potential Difference Created by Point Charges a x Example Where is the Potential equal to Zero? 0 V = Two point charges are fixed in place. The positive charge is +2q and the negative charge is –q. On the line that passes through the charges, how many places are there at which the total potential is zero? Where are they? kq1 kq2 q q + 2q −q + =0 → 1 + 2 =0 → + =0 r1 r2 r1 r2 |x| |x−a| x<0 2 1 2 1 → − =0 → = 3 cases : 0 < x < a |x| |x−a| |x| |x−a| x>a Extraneous root: x<0 amd x=2a are not compatible x < 0 → | x |= − x, | x − a |= a − x → 2( a − x ) = − x → 2a − 2 x = − x → x = 2a 0 < x < a → | x |= x, | x − a |= a − x → 2( a − x ) = x → 2a − 2 x = x → 3x = 2a → x = 2 a 3 x > a → | x |= x, | x − a |= x − a → 2( x − a ) = x → 2 x − 2a = x → x = 2a 9 19.5 Capacitors and Dielectrics A capacitor is a device designed to store energy and charge. E.g. defibrillators consist of a large capacitor that discharges current (flow of charges) when the paddles are applied to a patient In electrical circuits they often are used to stabilize electric potential Charge q (-q) is proportional to the potential difference between the plates. q = CV SI Unit of Capacitance: coulomb/volt = farad (F) 1 F=1 C/V=1 C2/J Memory Aid q C= V coulomb farad = volt France Britain : Italy +q −q 10 19.5 Capacitors and Dielectrics A THE CAPACITANCE OF A PARALLEL PLATE CAPACITOR Simplest case: air/vacuum gap (no “dielectric”) V = potential difference from the − to the + plate (by this definition, V for a capacitor is always positive) q = charge on the + plate (−q on the − plate) A = area of the plates d = plate separation/gap Note the different notation used by the textbook for this section Potential at distance y from negative plate; V ( y ) = Ey +q V d 0 Potential at positive plate is then qd V = Ed = E0 d = ε0 A −q A Here the subscript 0 indicates the special case of vacuum/air gap Electric field in vacuum/air (downward) : σ q = E = E0 = ε0 ε0 A C = C0 = qd ε 0 A q =q÷ = ε0 A d V Parallel plate capacitor filled with vacuum/air ε0 A C = C0 = d 11 19.5 Capacitors and Dielectrics A Example: A parallel plate capacitor have the following attributes: Area of plates: A=4.50m2 The gap is filled with air Plate separation: d=2.00cm + q Find its capacitance C d V -q A 12 19.5 Capacitors and Dielectrics A Example: A parallel plate capacitor have the following attributes: Area of plates: A=4.50m2 Plate separation: d=2.00cm The gap is filled with air Find its capacitance C +q d V Parallel plate capacitor filled with vacuum/air ε0 A C = C0 = d [8.85 × 10 −12 C 2 /( N ⋅ m 2 )]( 4.50 m 2 ) C = C0 = (0.0200 m) = 1.99 × 10 −9 F = 1.99 nF -q A Note that a farad (F) is a very large capacitance This 4.50 m2 plates at 2 .00 cm separation gives only ~2 nF 13 Capacitors as an energy storage device: It takes work (i.e. energy) to charge up a capacitor from zero charge to q (zero potential to V). +V + + + + +q + + + + ∆q 0 − − − − − −q − − − V=q/C V ∆q ∆W = (∆q)V q The figure shows a capacitor at charge q, potential difference V (between the − plate and the + plate). To increase q and V, we move a small amount of charge ∆q from the − plate to the + plate. This requires work done BY an external agent, AGAINST the electric field: ∆EPE = WEXT = −W− + (done by the electric field) = ( ∆q) Ed = ( ∆q)V = area of the shaded stripe in V vs. q graph 14 Capacitors as an energy storage device: (continued) +V + + + + +q + + + + ∆q 0 − − − − − −q − − − V=q/C V Energy = ∆W U = area of triangle ∆q = (∆q)V 2 =½ qV = ½ q /C q To charge a capacitor to (q, V) from (0,0), the total amount of work = area enclosed by the blue triangle, which is the energy stored in the capacitor. q2 Energy = U = 2C but q = CV Energy = U = 12 CV 2 15 Example: Parallel plate capacitor Area of plates: A=4.50m2 Plate charge: q=22.5µC Plate separation: d=2.00cm The gap is filled with air A +q (a) Find the voltage (potential difference) of the capacitor. (b) Find the energy stored in this capacitor d -q A 16 Example: Parallel plate capacitor A A=4.50m2 Area of plates: Plate separation: d=2.00cm q2 1 Plate charge: q=22.5µC Energy = U = = 2 CV 2 The gap is filled with air 2C +q d (a) Find the voltage (potential difference) of the capacitor. (b) Find the energy stored in this capacitor 22.5 × 10 −6 C q 22.5 × 10 −9 C (a) V = = = = 1.13 × 10 4 V 2 −9 −9 C 1.99 × 10 F 1.99 × 10 C / J (b) Energy = U = 12 CV 2 = 12 (1.99 × 10 −9 C 2 / J )(1.13 × 10 4 V) 2 -q A = 12 q 2 / C = 12 ( 22.5 × 10 −9 C) 2 /(1.99 × 10 −9 F) = 0.127 J Application: Typical defibrillators have banks of capacitors charged up to ~1000 V and usually an energy of up to ~200 J is delivered to the patient This means C = 2U/V2 ≈ 0.4 mF Resident Evil? So how do we make capacitance much bigger (approx. ×105)? 17 19.5 Capacitors and Dielectrics Dielectrics are materials that can be electrically polarized by the application of an external electric field. Eo = q ε0 A E= Eo Filling the gap of a capacitor with a dielectric: the molecules align such that “bound” charges (that are not free to move around as charges in a conductor) develop at the surface. The induced surface charges (opposite in sign to the charges on the plates) reduce the actual electric field inside the material. Note: dielectric = dia-electric; “dia” means “against” (Greek root) The factor by which the applied field E0 is reduced (to E), is called: THE DIELECTRIC CONSTANT Dielectric constant Eo κ= E Field in absence of dielectric Field in presence of dielectric κ 18 19.5 Capacitors and Dielectrics The Effect of a Dielectric When a Capacitor Has a Constant Charge Eo = q ε0 A E= Eo An empty capacitor is connected to a battery and charged up. The capacitor is then disconnected from the battery, and a slab of dielectric material is inserted between the plates. Does the voltage across the plates increase, remain the same, or decrease? Electric field in dielectric (κ > 1) : q is held constant → E is reduced E= E0 κ = σ q = κε 0 κε 0 A Potential difference between plates (dividing out q0 ) : V = ( ∆V ) − + = E ( ∆y ) − + = Ed = V is reduced by a factor of κ q κε A C= = 0 V d C increases κ - fold qd κε 0 A Parallel plate capacitor filled with dielectric κε 0 A C = κC0 = d κ 19 Example: Parallel plate capacitor in a defibrillator A (note this is both a high capacitance AND a precision application) Area of plates: A=0.0100m2 (4”x4”: fits in the enclosure) Maximum potential: V=1.00×103 V Maximum energy stored: U=200 J The gap is filled with BaTiO3 (barium titanate) : κ=5.00 ×103 +q d Find the required plate separation d and the capacitance C 2 κε κε A AV 2 0 0 U = 12 CV = V = 2d d κε 0 AV 2 →d = = 2U (5.00 × 10 3 )[8.85 × 10 −12 C 2 /( N ⋅ m 2 )](0.0100 m 2 )(1.00 × 10 3 J/C) 2 2( 200 J) 2 1 2 -q A d = 1.11 × 10 −6 m = 1.11 μm C= κε 0 A d (5.00 × 10 3 )[8.85 × 10 −12 C 2 /( N ⋅ m 2 )](0.0100 m 2 ) = 1.11x10 −6 C = 3.98 × 10 −4 F = 0.398 mF 20 19.4 Equipotential Surfaces and Their Relation to the Electric Field An equipotential surface is a surface on which the electric potential is the same everywhere. In this case they are for the point chareg potential kq V= r The net electric force does no work on a charge as it moves on an equipotential surface. The equipotential surfaces are analogous to the “contours” (curves on the boundaries between different colors instead of surfaces in this picture) on an elevation map such as this (V is analogous to height) 21 19.4 Equipotential Surfaces and Their Relation to the Electric Field The electric field created by any charge or group of charges is everywhere perpendicular to the associated equipotential surfaces and points in the direction of decreasing potential. Equipotential surfaces are typically drawn at equal intervals The electric field is always perpendicular to the surface pof a conductor: Conducting surfaces are equipotential surfaces! (the whole conductor is at the same potential) kq V= r 22