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Chapter 17 Electric Potential

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Chapter 17 Electric Potential Chapter 17 Electric Potential The Electric Potential Difference Created by Point Charges When two or more charges are present, the potential due to all of the charges is obtained by adding together the individual potentials. Example: The Total Electric Potential At locations A and B, find the total electric potential. The Electric Potential Difference Created by Point Charges 8.99×109 N ⋅m2 C2 + 8.0×10−-98 C 8.99×109 N ⋅m2 C2 −8.0×10−-98 C V = ( )( )+ ( )( )= +240 V A 0.20 m 0.60 m 8.99×109 N ⋅m2 C2 + 8.0×10−-98 C 8.99×109 N ⋅m2 C2 −8.0×10-9−8 C V = ( )( )+ ( )( )= 0 V B 0.40 m 0.40 m Equipotential Surfaces and Their Relation to the Electric Field An equipotential surface is a surface on which the electric potential is the same everywhere. Equipotential surfaces for kq a point charge (concentric V = r spheres centered on the charge) The net electric force does no work on a charge as it moves on an equipotential surface. 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. In equilibrium, conductors are always equipotential surfaces. Equipotential Surfaces and Their Relation to the Electric Field Lines of force and equipotential surfaces for a charge dipole. Equipotential Surfaces and Their Relation to the Electric Field The electric field can be shown to be related to the electric potential. If you move a distance Δs and the electric potential changes by an amount ΔV, the component of the electric field in the direction of Δs is the negative of the ratio of ΔV to Δs, or more compactly, ΔV E = − Δs ΔV/Δs is called the potential gradient. Equipotential Surfaces and Their Relation to the Electric Field Example: The Electric Field and Potential Are Related The plates of the capacitor are separated by a distance of 0.032 m, and the potential difference between them is VB-VA=-64V. Between the two equipotential surfaces shown in color, there is a potential difference of -3.0V. Find the spacing between the two colored surfaces. Equipotential Surfaces and Their Relation to the Electric Field ΔV −64 V 3 E = − = − = 2.0 ×10 V m Δs 0.032 m ΔV −3.0 V 3 Δs = − = − =1.5×10− m E 2.0 ×103 V m The gradient equation can be simplified for a parallel plate capacitor with voltage V across it and with plates separated by distance d as: V E = d Capacitors and Dielectrics A parallel plate capacitor consists of two metal plates, one carrying charge +q and the other carrying charge –q. It is common to fill the region between the plates with an electrically insulating substance called a dielectric. Advantages of dielectrics: • they break down (spark) less easily than air • the capacitor plates can be put closer together • they increase the ability to store charge Capacitors and Dielectrics THE RELATION BETWEEN CHARGE AND POTENTIAL DIFFERENCE FOR A CAPACITOR The magnitude of the charge in each place of the capacitor is directly proportional to the magnitude of the potential difference between the plates. q = CV The capacitance C is the proportionality constant. SI Unit of Capacitance: coulomb/volt = farad (F) The larger C is, the more charge the capacitor can hold for a given V Capacitors and Dielectrics E0 THE DIELECTRIC CONSTANT If a dielectric is inserted between the plates of a capacitor, the capacitance can increase markedly. E Dielectric constant κ = o E Since E ≤ E0 --> κ ≥ 1 Some lines of force now terminate or start on induced surface charges E ≤ E0 E0 E Capacitors and Dielectrics Capacitors and Dielectrics THE CAPACITANCE OF A PARALLEL PLATE CAPACITOR Eo = q (ε o A) E V E = o = κ d & κεo A # q = $ !V <--> q = CV % d " κε A Parallel plate capacitor C = o filled with a dielectric d C increases for --> increasing κ, increasing A, decreasing d Capacitors and Dielectrics Example: A Computer Keyboard One common kind of computer keyboard is based on the idea of capacitance. Each key is mounted on one end of a plunger, the other end being attached to a movable metal plate. The movable plate and the fixed plate form a capacitor. When the key is pressed, the capacitance increases. The change in capacitance is detected, thereby recognizing the key which has been pressed. The separation between the plates is 5.00 mm, but is reduced to 0.150 mm when a key is pressed. The plate area is 9.50x10-5m2 and the capacitor is filled with a material whose dielectric constant is 3.50. Determine the change in capacitance detected by the computer. Capacitors and Dielectrics −12 2 2 −5 2 κε o A (3.50)8.85×10 C N⋅m 9.50 ×10 m −12 C = = ( ( ))( ) =19.6×10 F d 0.150×10-3 m −12 2 2 −5 2 κε o A (3.50)8.85×10 C N⋅m 9.50 ×10 m −12 C = = ( ( ))( ) = 0.589×10 F d 5.00×10-3 m ΔC =19.0×10−12 F Example. How large would the plates have to be for a 1 F parallel plate capacitor with plate separation a) d=1 mm in air, b) d=1 µm separated by paper? a) d = 1 mm C = κε0A/d = ε0A/d (since κ = 1 for air) --> A = Cd/ε = (1)(.001)/(8.85 x 10-12) = 1.13 x 108 m2 0 If the plates were square, they would be of length and width (A)1/2 = (1.13 x 108)1/2 = 11 km --> very large!! b) d = 1µm = 10-6 m ~ the thickness of paper – use paper: κ = 3.3 -6 -12 4 2 --> A = Cd/(κε0) = (1)(10 )/(3.3 x 8.85 x 10 ) = 3.42 x 10 m and (A)1/2 = (3.42 x 104)1/2 = 185 m --> about 2 football fields in length, improving! Capacitors and Dielectrics Conceptual Example: The Effect of a Dielectric When a Capacitor Has a Constant Charge 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? Effect of a Dielectric When a Capacitor Has a Constant Charge -- continued Before inserting dielectric: q = C0 V0, C0 = ε0A/d After inserting dielectric: q = C V, C = κε0A/d = κC0 C0 V0 = q = C V = κC0 V --> V = V0/κ --> V < V0 for κ > 1 i.e. the voltage across the plates decreases when the dielectric is inserted if charge is held constant Capacitors and Dielectrics ENERGY STORAGE IN A CAPACITOR Capacitors store charge at some potential difference, therefore they store EPE, since by definition for a point charge q, EPE = qV . Your book shows that the energy stored in a capacitor can be written several ways: 1 1 q2 Energy = qV = CV 2 = 2 2 2C 2 Using Energy = 1/2 CV , C = κε0A/d, and V = Ed , the energy of a capacitor can be written as, Volume = Ad 1 & κεo A # 2 Energy = 2 $ !(Ed ) % d " We can then calculate the energy density of a capacitor as, Energy 1 2 (valid in general for the Energy density = Volume = 2 κεo E energy density stored in an electric field) Example. How much energy is stored in a 10 F capacitor with a potential difference of 3 V across it? 1 1 2 Energy = CV 2 = (10)(3) = 45J 2 2 This is the same amount of energy stored in a 4.6 kg mass suspended 1 m above the ground. .
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