General Physics (PHY 2140) Lecture 5 Electrostatics Electrical energy potential difference and electric potential potential energy of charged conductors Capacitance and capacitors http://www.physics.wayne.edu/~apetrov/phy2140/ Chapter 16 1
Lightning Review Last lecture: 1. Flux. Gauss s law. simplifies computation of electric fields Φ = EAcosθ = net Q ε o 2. Potential and potential energy V = VB VA = electrostatic force is conservative potential (a scalar) can be introduced as potential energy of electrostatic field per unit charge PE q Review Problem: Perhaps you have noticed sudden gushes of rain or hail moments after lightning strokes in thunderstorms. Is there any connection between the gush and the stroke or thunder? Or is this just a coincidence? V cloud E F = C mg qe 2
16.2 Electric potential and potential energy due to point charges Electric circuits: point of zero potential is defined by grounding some point in the circuit Electric potential due to a point charge at a point in space: point of zero potential is taken at an infinite distance from the charge With this choice, a potential can be found as V = q ke r Note: the potential depends only on charge of an object, q,, and a distance from this object to a point in space, r.. 3
Superposition principle for potentials If more than one point charge is present, their electric potential can be found by applying superposition principle The total electric potential at some point P due to several point charges is the algebraic sum of the electric potentials due to the individual charges. Remember that potentials are scalar quantities! 4
Potential energy of a system of point charges Consider a system of two particles If V 1 is the electric potential due to charge q 1 at a point P, then work required to bring the charge q 2 from infinity to P without acceleration is q 2 V 1. If a distance between P and q 1 is r, then by definition q 2 q 1 P r A PE = q V = k 2 1 e qq r 1 2 Potential energy is positive if charges are of the same sign and vice versa. 5
Mini-quiz: potential energy of an ion Three ions, Na +, Na +, and Cl -, located such, that they form corners of an equilateral triangle of side 2 nm in water. What is the electric potential energy of one of the Na + ions?? Cl - Na + Na + q q q q q PE = k + k = k q + q r r r [ ] Na Cl Na Na Na e e e Cl Na but : q = q! Cl Na qna PE = ke qna + qna = r [ ] 0 6
16.3 Potentials and charged conductors Recall that work is opposite of the change in potential energy, [ ] W = PE = q V V No work is required to move a charge between two points that are at the same potential. That is, W=0 if V B =V A Recall: 1. all charge of the charged conductor is located on its surface 2. electric field, E, is always perpendicular to its surface, i.e. no work is done if charges are moved along the surface Thus: potential is constant everywhere on the surface of a charged conductor in equilibrium B A but that s not all! 7
Because the electric field is zero inside the conductor, no work is required to move charges between any two points, i.e. [ ] 0 W = q V V = B If work is zero, any two points inside the conductor have the same potential, i.e. potential is constant everywhere inside a conductor Finally, since one of the points can be arbitrarily close to the surface of the conductor, the electric potential is constant everywhere inside a conductor and equal to its value at the surface! A Note that the potential inside a conductor is not necessarily zero, even though the interior electric field is always zero! 8
The electron volt A unit of energy commonly used in atomic, nuclear and particle physics is electron volt (ev( ev) The electron volt is defined as the energy that electron (or proton) gains when accelerating through a potential difference of 1 V Relation to SI: V ab =1 V 1 ev = 1.60 10 10-19 C V V = 1.60 10 10-19 J 9
Problem-solving strategy Remember that potential is a scalar quantity Superposition principle is an algebraic sum of potentials due to a system of charges Signs are important Just in mechanics, only changes in electric potential are significant, hence, the point you choose for zero electric potential is arbitrary. 10
Example : ionization energy of the electron in a hydrogen atom In the Bohr model of a hydrogen atom, the electron, if it is in the ground state, orbits the proton at a distance of r = 5.29 10-11 m. Find the ionization energy of the atom, i.e. the energy required to remove the electron from the atom. Note that the Bohr model, the idea of electrons as tiny balls orbiting the nucleus, is not a very good model of the atom. A better picture is one in which the electron is spread out around the nucleus in a cloud of varying density; however, the Bohr model does give the right answer for the ionization energy 11
In the Bohr model of a hydrogen atom, the electron, if it is in the ground state, orbits the proton at a distance of r = 5.29 x 10-11 m. Find the ionization energy, i.e. the energy required to remove the electron from the atom. Given: r = 5.292 x 10-11 m m e = 9.11 10-31 kg m p = 1.67 10-27 kg e = 1.60 10-19 C Find: E=? The ionization energy equals to the total energy of the electron-proton system, E = PE + KE with e PE = ke, KE = m r v 2 2 2 The velocity of e can be found by analyzing the force on the electron. This force is the Coulomb force; because the electron travels in a circular orbit, the acceleration will be the centripetal acceleration: ma = F or e c c m Thus, total energy is e v r e = k, e 2 or r 2 2 v 2 = e e 2 ke mr e, 2 2 2 e m e kee e E = ke + = ke = r 2 mer 2r 18 2.18 10 J -13.6 ev 12
16.4 Equipotential surfaces They are defined as a surface in space on which the potential is the same for every point (surfaces of constant voltage) The electric field at every point of an equipotential surface is perpendicular to the surface convenient to represent by drawing equipotential lines 13
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16.6 The definition of capacitance Capacitor: two conductors (separated by an insulator) usually oppositely charged a +Q b -Q The capacitance, C, of a capacitor is defined as a ratio of the magnitude of a charge on either conductor to the magnitude of the potential difference between the conductors Q C = V 15
1. A capacitor is basically two parallel conducting plates with insulating material in between. The capacitor doesn t have to look like metal plates. 2. When a capacitor is connected to an external potential, charges flow onto the plates and create a potential difference between the plates. 3. Capacitors in circuits symbols analysis follow from conservation of energy conservation of charge Capacitor for use in high-performance audio systems. - - + - 16
Units of capacitance The unit of C is the farad (F), but most capacitors have values of C ranging from picofarads to microfarads (pf to µf). 1F = 1CV Recall, micro 10-6, nano 10-9, pico 10-12 If the external potential is disconnected, charges remain on the plates, so capacitors are good for storing charge (and energy). 17
16.7 The parallel-plate plate capacitor The capacitance of a device depends on the geometric arrangement of the conductors C = ε 0 A d d +Q -Q A A where A is the area of one of the plates, d is the separation, ε 0 is a constant called the permittivity of free space, k e = 1 4πε 0 ε 0 = 8.85 10-12 C 2 /N m2 18
Problem: parallel-plate plate capacitor A parallel plate capacitor has plates 2.00 m 2 in area, separated by a distance of 5.00 mm. A potential difference of 10,000 V is applied across the capacitor. Determine the capacitance the charge on each plate 19
A parallel plate capacitor has plates 2.00 m 2 in area, separated by a distance of 5.00 mm. A potential difference of 10,000 V is applied across the capacitor. Determine the capacitance the charge on each plate Solution: Given: V=10,000 V A = 2.00 m 2 d = 5.00 mm Find: C=? Q=? Since we are dealing with the parallel-plate capacitor, the capacitance can be found as A C = ε = C N m d 9 = 3.54 10 F = 3.54 nf 12 2 2 ( 8.85 10 ) 2.00 m 5.00 10 0 3 Once the capacitance is known, the charge can be found from the definition of a capacitance via charge and potential difference: ( )( ) 9 5 Q= C V = 3.54 10 F 10000V = 3.54 10 C 2 m 20
16.8 Combinations of capacitors It is very often that more than one capacitor is used in an electric circuit We would have to learn how to compute the equivalent capacitance of certain combinations of capacitors C 1 C 2 C 1 C 5 C 3 C 3 C 2 C 4 21
a. Parallel combination a Connecting a battery to the parallel combination of capacitors is equivalent to introducing the same potential difference for both capacitors, V=V ab b By definition, C 1 +Q 1 C 2 +Q 2 Q 1 Thus, C eq would be Q 2 Q = CV Q = C V C 1 1 1 2 2 2 eq eq eq 11 22 V1 = V2 = V A total charge transferred to the system from the battery is the sum of charges of the two capacitors, Q1+ Q2 = Q Q Q + Q Q Q Q Q = = + = + V V V V V V C = C + C 1 2 1 2 1 2 Q + Q = Q 1 2 V = V = V 1 2 1 2 22
Parallel combination: notes Analogous formula is true for any number of capacitors, C = C + C + C + (parallel combination) eq 1 2 3... It follows that the equivalent capacitance of a parallel combination of capacitors is greater than any of the individual capacitors 23
Problem: parallel combination of capacitors A 3 µf capacitor and a 6 µf capacitor are connected in parallel across an 18 V battery. Determine the equivalent capacitance and total charge deposited. 24
A 3 µf capacitor and a 6 µf capacitor are connected in parallel across an 18 V battery. Determine the equivalent capacitance and total charge deposited. a Given: C 1 +Q 1 C 2 +Q 2 V = 18 V C 1 = 3 µf C 2 = 6 µf V=V ab b Q 1 Q 2 Find: C eq =? Q=? First determine equivalent capacitance of C 1 and C 2 : C C C µ F 12 = 1 + 2 = 9 Next, determine the charge ( )( ) 6 4 Q= C V = 9 10 F 18V = 1.6 10 C 25
b. Series combination V=V ab b a By definition, C 1 C 2 Thus, C eq would be c +Q 1 Q 1 +Q 2 Q 2 Q = CV Q = CV 1 1 1 2 2 2 1 1 2 2 Connecting a battery to the serial combination of capacitors is equivalent to introducing the same charge for both capacitors, Q1 = Q2 = Q A voltage induced in the system from the battery is the sum of potential differences across the individual capacitors, 1 2 1 2 V = V1+ V2 1 V V1+ V2 V1 V2 V1 V2 = = + = + C Q Q Q Q Q Q eq 1 1 1 = + + C C C 1 1 1 C C C eq eq Q = Q = Q V + V = V 1 2 26
Series combination: notes Analogous formula is true for any number of capacitors, 1 1 1 1 = + + +... (series combination) C C C C eq 1 2 3 It follows that the equivalent capacitance of a series combination of capacitors is always less than any of the individual capacitance in the combination 27
Problem: series combination of capacitors A 3 µf capacitor and a 6 µf capacitor are connected in series across an 18 V battery. Determine the equivalent capacitance. 28
A 3 µf capacitor and a 6 µf capacitor are connected in series across an 18 V battery. Determine the equivalent capacitance and total charge deposited. a Given: V = 18 V C 1 = 3 µf C 2 = 6 µf V=V ab b C 1 C 2 c +Q 1 Q 1 +Q 2 Q 2 Find: C eq =? Q=? First determine equivalent capacitance of C 1 and C 2 : C eq CC C + C 1 2 = = 1 2 Next, determine the charge 2 µ F ( )( ) 6 5 Q= C V = 2 10 F 18V = 3.6 10 C 29