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Ch. 20 Electric Potential and Electric Potential Energy

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Electric Potential Energy Electrical potential energy is the energy contained in a configuration of charges. Like all potential energies, when it goes up the configuration is less stable; when it goes down, the configuration is more stable. The unit is the Joule.

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Equation U = - W = q 0 Ed q 0 – test charge E – Electric field d - distance

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Electrical potential energy increases when charges are brought into less favorable configurations

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Electrical potential energy decreases when charges are brought into more favorable configurations.

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Work must be done on the charge to increase the electric potential energy

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For a positive test charge to be moved upward a distance d, the electric force does negative work. The electric potential energy has increased and U is positive (U2 > U1)

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If a negative charge is moved upward a distance d, the electric force does positive work. The change in the electric potential energy U is negative (U2 < U1)

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Electric Potential (V) Electric potential is hard to understand, but easy to measure. We commonly call it voltage, and its unit is the Volt. 1 V = 1 J/C Electric potential is easily related to both the electric potential energy, and to the electric field.

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The change in potential energy is directly related to the change in voltage. U = q V V = U/q U: change in electrical potential energy (J) q: charge moved (C) V: potential difference (V) All charges will spontaneously go to lower potential energies if they are allowed to move.

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Since all charges try to decrease UE, and U E = q V, this means that spontaneous movement of charges result in negative U. V = U / q Positive charges like to DECREASE their potential ( V < 0) Negative charges like to INCREASE their potential. ( V > 0)

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Sample Problem: A 3.0 μC charge is moved through a potential difference of 640 V. What is its potential energy change?

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Electrical Potential in Uniform Electric Fields The electric potential is related in a simple way to a uniform electric field. V = -Ed V: change in electrical potential (V) E: Constant electric field strength (N/C or V/m) d: distance moved (m)

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Sample Problem: An electric field is parallel to the x-axis. What is its magnitude and direction if the potential difference between x =1.0 m and x = 2.5 m is found to be +900 V?

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Sample Problem: If a proton is accelerated through a potential difference of V, what is its change in potential energy? How fast will this proton be moving if it started at rest?

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Sample Problem: A proton at rest is released in a uniform electric field. What potential difference must it move through in order to acquire a speed of 0.20 c?

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Electric Potential Energy for Spherical Charges Electric potential energy is a scalar, like all forms of energy. U = kq 1 q 2 /r U: electrical potential energy (J) k: 8.99 × 10 9 N m 2 / C 2 q 1, q 2 : charges (C) r: distance between centers (m) This formula only works for spherical charges or point charges.

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Absolute Electric Potential (spherical) For a spherical or point charge, the electric potential can be calculated by the following Formula: V = kq/r V: potential (V) k: 8.99 x 10 9 N m 2 /C 2 q: charge (C) r: distance from the charge (m) Remember, k = 1/(4 o )

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Electric Field and Electric Potential E = - V / d Two things about E and V: The electric field points in the direction of decreasing electric potential. The electric field is always perpendicular to the equipotential surface.

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20-4 Equipotential Surfaces and the Electric Field For two point charges:

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Equipotential Surfaces and the Electric Field An ideal conductor is an equipotential surface. Therefore, if two conductors are at the same potential, the one that is more curved will have a larger electric field around it. This is also true for different parts of the same conductor.

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Equipotential Surfaces and the Electric Field There are electric fields inside the human body; the body is not a perfect conductor, so there are also potential differences. An electrocardiograph plots the hearts electrical activity.

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Equipotential Surfaces and the Electric Field An electroencephalograph measures the electrical activity of the brain:

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Capacitor Named for the capacity to store electric charge and energy. A capacitor is two conducting plates separated by a finite distance:

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The capacitance relates the charge to the potential difference:

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Sample Problem: A 0.75 F capacitor is charged to a voltage of 16 volts. What is the magnitude of the charge on each plate of the capacitor?

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V = 16 V, C = 0.75 F = 0.75 x F Q =? C = Q/V or Q = CV Q = (0.75 x )(16) Q = 1.2 x C

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A simple type of capacitor is the parallel-plate capacitor. It consists of two plates of area A separated by a distance d. By calculating the electric field created by the charges ±Q, we find that the capacitance of a parallel-plate capacitor is:

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The general properties of a parallel-plate capacitor – that the capacitance increases as the plates become larger and decreases as the separation increases – are common to all capacitors.

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Capacitor Geometry The capacitance of a capacitor depends on HOW you make it.

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Sample Problem: What is the AREA of a 1 F capacitor that has a plate separation of 1 mm? C = 1 F, d = 1 mm = m, = 8.85 x C 2 /(Nm 2 )

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1.13 x 10 8 m m Is this a practical capacitor to build? NO! – How can you build this then? The answer lies in REDUCING the AREA. But you must have a CAPACITANCE of 1 F. How can you keep the capacitance at 1 F and reduce the Area at the same time? Add a DIELECTRIC!!!

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A dielectric is an insulator; when placed between the plates of a capacitor it gives a lower potential difference with the same charge, due to the polarization of the material. This increases the capacitance.

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Dielectric Remember, the dielectric is an insulating material placed between the conductors to help store the charge. In the previous example we assumed there was NO dielectric and thus a vacuum between the plates. All insulating materials have a dielectric constant associated with it. Here now you can reduce the AREA and use a LARGE dielectric to establish the capacitance at 1 F.

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Sample Problem: A parallel plate capacitor is constructed with plate of an area of m 2 and a separation of 0.55 mm. Find the magnitude of the charge of this capacitor when the potential difference between the plate is 20.1 V. A = m 2 d = 0.55 mm = m, V = 20.1 V Q = ? Q = CV C = A/d C=(8.85 x )(0.028) /( ) C = 4.51 x F Q = (4.51 x )(20.1) Q = 9.06 x C

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The polarization of the dielectric results in a lower electric field within it; the new field is given by dividing the original field by the dielectric constant κ: Therefore, the capacitance becomes:

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The dielectric constant is a property of the material; here are some examples:

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Using MORE than 1 capacitor Lets say you decide that 1 capacitor will not be enough to build what you need to build. You may need to use more than 1. There are 2 basic ways to assemble them together Series – One after another Parallel – between a set of junctions and parallel to each other.

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Capacitors in Series Capacitors in series each charge each other by INDUCTION. So they each have the SAME charge. The electric potential on the other hand is divided up amongst them. In other words, the sum of the individual voltages will equal the total voltage of the battery or power source.

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Capacitors in Parallel In a parallel configuration, the voltage is the same because ALL THREE capacitors touch BOTH ends of the battery. As a result, they split up the charge amongst them.

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Capacitors STORE energy Anytime you have a situation where energy is STORED it is called POTENTIAL. In this case we have capacitor potential energy, U c Suppose we plot a V vs. Q graph. If we wanted to find the AREA we would MULTIPLY the 2 variables according to the equation for Area. A = bh When we do this we get Area = VQ Lets do a unit check! Voltage = Joules/Coulomb Charge = Coulombs Area = ENERGY

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Potential Energy of a Capacitor Since the AREA under the line is a triangle, the ENERGY(area) =1/2VQ This energy or area is referred as the potential energy stored inside a capacitor. Note: The slope of the line is the inverse of the capacitance. most common form

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Sample Problem: In a typical defibrillator, a 175 F, is charged until the potential difference between the plates is 2240 V. A.) What is the charge on each plate? V = 2240 V, C = 175 F = 175 x F Q = ? Q = CV Q = (175 x )(2240) Q = C

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B.) Find the energy stored in the charged up defibrillator. U = ? Since you now know Q, C, & V. You may use any of the 3 equations to find U. U = ½ CV 2 U = ½ QV U = Q 2 /(2C) U = ½ (0.392)(2240) U = 439 J

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The energy stored in a capacitor can be put to a number of uses: a camera flash; a cardiac defibrillator; and others. In addition, capacitors form an essential part of most electrical devices used today. If we divide the stored energy by the volume of the capacitor, we find the energy per unit volume; this result is valid for any electric field:

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If the electric field in a dielectric becomes too large, it can tear the electrons off the atoms, thereby enabling the material to conduct. This is called dielectric breakdown; the field at which this happens is called the dielectric strength.

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