* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Book 1
Survey
Document related concepts
Speed of gravity wikipedia , lookup
Casimir effect wikipedia , lookup
Work (physics) wikipedia , lookup
History of electromagnetic theory wikipedia , lookup
Electrical resistivity and conductivity wikipedia , lookup
Potential energy wikipedia , lookup
Introduction to gauge theory wikipedia , lookup
Maxwell's equations wikipedia , lookup
Electromagnetism wikipedia , lookup
Anti-gravity wikipedia , lookup
Field (physics) wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Lorentz force wikipedia , lookup
Transcript
PHYSICS 126 Lecture Notes Book 1 Prepared by Kai Wong Table of Contents 1. Charging Processes ………………………………………………………… 2 2. Coulomb’s Law …………………………………………………………. 7 3. Electric Field ……………………………………………………. 11 4. Electric Potential I …………………………………………. 19 5. Electric Potential II ………………………………………………………… 27 6. Capacitance …………………………………………………………………… 33 1 1. Charging Processes Atomic Origin of Electricity Atoms are composed of electrons and nuclei, which permanently carry negative and positive electric charges respectively. The charge carried by a single electron is equal to e , where e 1.6 1019 C The symbol C stands for Coulomb, which is the SI unit for charge. A basic property of electric charges is that like-charges repel and unlike-charges attract. The force weakens with distances. The atom is neutral as a whole, carrying as many units of positive as negative electric charges. It becomes an ion when it loses or gains electrons. Aggregate matter is therefore usually neutral, and becomes noticeably charged only if a significant amount of charge of one kind has been transferred into or out of the matter, in the form of transfer of electrons or ions. Example: How much is the charge left on 1 cc of water after all the electrons have been removed? Solution: Since the density of water is 1 g/cm3, the mass is 1g. From the molecular formula H2O for water, the molecular weight is 2 1 16 18 . 1 number of moles 0.056 18 number of molecules 0.056 6.022 1023 3.37 1022 The number of electrons per molecule = 2+8=10 electric ch arg e 3.37 1022 10 1.6 1019 C 5.4 104 C When a glass rod is rubbed with silk cloth, electrons are transferred from the rod to the cloth. As a result, the glass rod becomes positively charged, while the silk cloth becomes negatively charged. On the other hand, when a plastic rod is rubbed with fur, the charge transferred is such that the plastic rod becomes negatively charged, while the fur is positively charged. Because charging arises from separation and transfer of charges, no net charge is created in the universe by the charging process. This fact is called the conservation of electric charge, and can be expressed as Q Q after becore 2 where the summation is over all objects involved in the charge transfer process, and the charge Q carries the proper sign. Ordinary matter can be classified into conductors and insulators. All metals are conductors. In a conductor, a large number of electrons are ionised from the atoms, and are called free electrons, because they can move freely inside the conductor in response to electric forces. Insulators are such materials as plastics, dried wood, and glasses. All electrons in an insulator are bound to atomic nuclei , so that charges cannot flow freely. If a conductor is isolated and contains an excess of electrons so that it becomes negatively charged, the excess electrons tend to move to populate the surface of the conductor because of their mutual repulsion. Likewise, if the conductor is depleted in electrons so that it is positively charged as a whole, the excess positive charges also reside on the surface. Excess charge on an isolated conductor stays on its surface: _ _ _ _ _ _ _ _ _ _ _ _ Charging by Contact When two conductors are brought into contact, they behave as a single conductor so that electrons can flow freely among the two. In particular, a neutral conductor can be charged up after contact with a charged conductor, and bears charge of the same kind. The earth itself is a giant conductor. If a charged conductor is brought in contact with the earth, as happens when it is touched by your hands and you are not wearing rubber shoes, the charge on the conductor will be shared with the earth. Since the latter is so much bigger, almost all the charge on the conductor is drained to the earth. This phenomenon is known as grounding. To avoid grounding, the conductor should stand on an insulator to isolate from the earth. 3 + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + Example: Two identical metal spheres initially contain charges 5q and 9q respectively. They are brought into contact and then separated. What are the charges on them after separation? Solution: Since the two spheres are identical, the charges are equally distributed between while in contact. Let this charge be Q . From charge conservation: 2Q 5q 9q Q 2q Charging by Induction When a positively charged object is brought near a neutral conductor but not in contact, the free electrons in the conductor are attracted by the object and rush to reside on the part of the surface near the object, leaving behind a net positive charge on the part of the 4 surface away from the object. A polarization is said to be induced on the conductor. When the object is removed, the free electrons will move to nullify the polarization so that the conductor is restored to its original configuration with no charge on the surface. On the other hand, if you touch the far end of the conductor while the positively charged object is still in place, the positive charge on the far end is conducted away to ground. If the object is now removed, there is a net negative charge left on the conductor, which, because of mutual repulsion, will spread out to occupy the surface. The isolated conductor is now negatively charged. + + + + + + + + + + + + + + _ __ _ + + + + + + + + + + + + _ __ _ + + + + + + + + + + _ _ _ _ _ + + + _ _ + _ _ _ _ _ _ - _ _ Polarization also occurs when a charged object is brought near certain insulators. These are insulators for which the charge distribution inside a molecule is skewed, so that one end of the molecule is slightly more positive while the other end is slightly more negative. The molecule is said to form an electric dipole. Ordinarily, the orientation of 5 these dipoles are random, so that in any given volume of the material, the net charge is zero. If a positively charged object is brought near such an insulator, the negative ends of the dipoles are attracted to the object, resulting in a lining up of the dipoles. Inside any given volume of the insulator, the net charge is still zero, because of cancellation between positive and negative charges. However, on the surface facing the object, the negative ends of the dipoles are not cancelled by positive charges from other dipoles. The net charge on this part of the surface is therefore negative. Similarly, the far end of the insulator becomes positively charged. Because the negative charge on the insulator is closer to the positively charged object, the insulator is attracted to the object with a stronger force than the repulsive force on its positive far end. If the insulator is light, it can be levitated and stick to the object. This is why shreds of paper can be picked up by a comb that you have rubbed against your hair. - + + - + + - + - + + - + - + - + - + - + - 6 + 2. Coulomb’s Law Two point charges at a distance r apart and carrying charges q1 and q2 respectively will either repel or attract each other with a force whose magnitude is given by F k q1 q2 r2 The force is an attraction if q1 and q2 have opposite signs, otherwise it is a repulsion. If the charges q1 and q2 are measured in C, the distance r is in meter, and the force F is in Newton, the constant k , known as the Coulomb’s law constant, is given by k 9.0 109 N m2 / C 2 Note that the forces on the two charges are equal and opposite, and form an action and reaction pair in the sense of Newton’s third law of motion. The force decreases with distance, being inversely proportional to the distance squared, similar to the gravitational force between two point masses. For convenience, we also introduce the constant 0 known as the permitivity of free space through the definition k 1 4 0 0 8.85 10 C 2 / N m 2 12 Example: Find the force between point charges 3nC and 2nC at a distance of 2m apart. Solution: The force is repulsive, with magnitude 3 109 2 109 F 9.0 109 13.5 10 9 N 2 2 7 Example: In the Bohr model of hydrogen atom, the electron is at a distance of r 5.29 1011 m from the nucleus. Find the electrostatic force of attraction on the electron due to the nucleus. Compare it with the gravitational force, also due to the nucleus. Solution: The electrostatic force is Felectric ke2 9.0 109 1.6 1019 2 2 r 5.26 1011 2 8.2 108 N Since Fgravity Gme m p r2 where the gravitational constant G 6.67 1011 N m2 / kg2 the mass of electron me 9.1 1031 kg mass of proton mp 1.67 1027 kg the ratio Fgravity Felectric Gme m p ke 2 4.4 10 41 so that the gravitation force is very much weaker. Example: Assuming the orbit of the electron in the previous example is a circle, find the speed of the electron. Solution: The electrostatic force on the electron always point toward the proton, and acts as the centripetal force for the circular motion. Therefore Felectric v me v 2 r Felectricr 2.18 10 6 m / s me 8 The proton experiences the same force Felectric directed toward the electron. However, its acceleration is negligible compared with that of the electron, because the proton mass is much larger. When more than two point charges are present, to calculate the force on one of the charges, it is necessary to add up the forces due to the rest of the charges. The addition of forces obeys the law of vector addition. Example: Three point charges qa 3C, qb 2C, qc 1C lie on a straight line. The distance between q a and q b is 2m. Find the force on q c if (1) it is closer to q b and at 1m from it. (2) It is midway between q a and q b Solution: (1) The configuration of charges is as shown: a 3μC b c 2μC 1μC 2m 1m The colored arrows represent the forces on q c due to q a (blue)and q b (red). We see that the former is a repulsion and the latter is an attraction. These forces are a on c: Fac 9.0 10 9 3 10 6 1 10 6 3 10 3 N 2 3 directed to the right b on c: Fbc 9.0 10 9 2 10 6 1 10 6 18 10 3 N 12 directed to the left The net force on q c is therefore 15 10 3 N directed to the left. (2) The charge configuration is c a b 3μC 1m 1μC 2μC 2m 9 The forces on q c due to q a (blue) and q b (red) are 3 10 6 1 10 6 Fac 9.0 10 27 10 3 N 2 1 to the right 2 10 6 1 10 6 18 10 3 N 2 1 to the right 9 Fbc 9.0 10 9 The net force is therefore 45 10 3 N to the right. 10 3. Electric Field Any collection of charges is called a charge distribution. A charge distribution can be discrete, consisting of a number of point charges, or it can be continuous, occupying a portion of a surface (2-D) or a volume (3-D) of space. A point charge introduced into the vicinity of a charge distribution is called a test charge. Let the charge it carries be q 0 . At any position the test charge is placed, it will experience a force F due to the charges in the distribution, and this force is proportional to q 0 according to Coulomb’s law. The quantity F E q0 is therefore independent of the charge q 0 , depending only on the charge distribution and the location of the point where the test charge is place. It is called the electric field at that location. It is equal to the force acting on a test charge of 1C . The unit of electric field is N/C (newton per coulomb). We can also write F q0 E which tells us how to calculate the force on a point charge at a location where the electric field is known. In particular, the direction of the force is the same as the direction of the electric field if the charge is positive, and is opposite if the charge is negative. Electric Field due to a point charge. The simplest charge distribution is a point charge q . It creates an electric field everywhere in space. At a point at a distance r from the charge, the magnitude of the electric field is, from Coulomb’s law, q E k 2 R The direction of E is radially outward from the point charge if it is positive, and inward toward the point charge if it is negative. The following figure displays the electric field vectors in a number of locations surrounding a positive and a negative charge respectively. 11 Electric Lines of Force These are imaginary lines with assigned arrows drawn in space with the property that at any point on the line, the direction of the electric field is tangent to the line and agrees with the direction of the arrow. A number of lines of force are drawn for the positive and point charge as shown: For a charge distribution consisting of more than one point charge or an arbitrary continuous distribution, the lines of force are curves. Customarily, we draw equally spaced lines of force emanating from a positive point charge or converging onto a negative point charge. The number of lines coming out or going into a point charge is proportional to the magnitude of the charge. The lines of force have the following properties: 1. Two different lines cannot intersect each other. (Because if they do, there would be two directions for the electric field at the point of intersection.) 2. The lines are not closed. They either come out of positive charges, terminate at negative charges, or extend to infinity. 3. Regions where the lines are more crowded together have stronger electric field. 12 Electric field due to a number of point charges The electric field due to a discrete point charge distribution can be calculated as the vector sum of the electric field due to the individual charges. One of the simplest discrete distribution is an electric dipole, consisting of a pair of point charges of equal in magnitude but opposite in sign. As an example, consider a dipole made up of charges 5nC and 5nC separated by a distance of 2m. Let’s find the electric field at the location A, which is at 1m from the negative charge as shown: 5nC 5nC A The electric field at A due to the 5nC charge (blue) is 9.0 10 9 5 10 9 5 N / C to the right (blue arrow) 32 The electric field at A due to the 5nC charge (red) is 5 10 9 9 9.0 10 45 N / C to the left (red arrow) 12 Therefore the electric field of the dipole at the point A is 40N/C and points to the left, and is indicated as a black arrow. Next, we’ll find the electric field at the point B as shown: y B θ x 2 5m m θ 5nC 1 m 5nC 13 The electric fields due to the blue and red charges at B are indicated as blue and red arrows. Since the distance between B and either one of the charge is 12 2 2 5 m, The fields have the same magnitude of 9.0 10 9 5 10 9 5 2 9N / C To add the blue and red vectors, we introduce x and y axes at the point B as shown. We have: sum of y components of the two vectors =0; sum of x components is 9 cos 9 cos 18 cos , where the angle θ can be calculated from the triangle on the left as cos 1 5 0.447 Therefore the electric field at the point B has magnitude 18 0.447 8.0N / C . Its direction is indicated by the black arrow. The following figure shows the field lines from a dipole: For two equal point charges of the same sign, the pattern of field lines is as shown: 14 Example: Four point charges lie on the corners of a square with side of length a as shown. Find the electric field at the center of the square. A B +q -q O -q -q D C Solution: At the center O, the electric fields due to the charges at the corners B and D cancel out because they are equal but in the opposite directions. On the other hand, the fields due to A and C are equal but in the same direction. The length AO, denoted by r , can be found from Pythagoras theorem: r 2 r 2 a2 r a 2 The magnitude of the electric field at O is therefore 2k q 4kq 2 r2 a and points toward the corner C. Example: Three point charges are at the corners of a square of side 1m as shown. Find the electric field at the unoccupied corner. A B 1m +4nC -4nC n C y x -4nC -q D C 15 Solution: In the figure above, the directions of the electric fields due to the three charges at the point D are colored. Their magnitudes are 4 10 9 36 N / C 12 E A EC 9.0 10 9 E B 9.0 10 9 4 10 9 2 2 18 N / C since the length BD is 2 from Pythagoras theorem. Choosing x and y axes at the point D as shown, we have, for the components of the net electric field, E x 18 cos 45 36 48.7 E y 18 sin 45 36 23.3 2 The magnitude of the electric field is E 48.7 2 23.3 54.0 N / C The angle it makes with the x-axis is tan 1 23.3 25.6 48.7 Electric field of an infinite charge sheet Consider an infinite plane sheet carrying a uniform 2-D distribution of charge. Let the surface charge density be . The unit of is C/m2. The pattern of field lines are shown below for positively charged and negatively charged sheets respectively. Positive sheet Negative sheet At any point on either side of the sheet, the magnitude of the electric field is E 2 0 16 For a pair of sheets bearing surface charge densities of equal magnitude but opposite signs, the magnitude of the electric field between the sheets is E 0 being twice as much as that due to a single sheet. Elsewhere the electric field is zero. This is because between the sheets, the contributions from the two sheets add, but they cancel elsewhere as shown in the figure( positive sheet and its electric field is in blue, negative sheet and its electric field in red) In practice, sheet charges are finite in extent. The formulas for infinite sheets are applicable at locations close to the surface compared with the extent. The configuration of parallel sheet charges of opposite signs can be created by connecting two metallic plates to the two terminals of a battery. This constitutes a parallel plate capacitor. Note that at the ends of a parallel plate capacitor, the field lines are curved. Well inside the plates, the electric field is uniform in space, meaning that it is independent of the location. Example: (Motion of electron in region of uniform electric field) An electron travelling horizontally at 5.0 10 7 m / s enters into the space between two parallel plates where the electric field is equal to 2.5 10 4 N / C and points downward. How large is the upward deflection of the electron after travelling 0.2m in the region? 17 Solution: y x E The force on the electron is upward and has magnitude F eE F eE y component of acceleration a y me me x component of acceleration a x 0 If v denotes the horizontal velocity when the electron enters the field region, the time to x travel a horizontal distance x is t v During this time, the vertical distance travelled by the electron is 1 1 eE x eEx 2 1.6 10 19 2.5 10 4 0.2 2 y ayt 2 0.035m 2 2 2 me v 2me v 2 2 9.1 10 31 5.0 10 7 2 Shielding Electrostatics deals with situations in which charges are not moving. In this case, the electric field inside a conductor is zero, because otherwise the free electrons will move. Excess charges in a conductor therefore reside on the surface in electrostatics. Since the electric field arises from the surface charges and other charges outside the conductor, the various contributions must cancel out inside the conductor. If we hollow out a volume inside the conductor, the surface charges and the charges outside the conductor are not disturbed. Therefore the electric field in the hollowing remains zero. This is obtained even if the hollowing is complete so that only a thin shell of the conductor remains. We thus arrive at a situation in which no matter how large the electric field is outside a conductor shell, there is no electric field inside the shell. This works even if the shell is replaced by a cage made with metallic wires. The phenomenon is known as shielding. + ++ + ++ + - E= 0 + + + For this reason, it is quite safe to sit in your car during thunderstorm. 18 4. Electric Potential Definition of Electric Potential Consider first the simple case of a region of space where the electric field E is uniform. In this region, a test charge q 0 experiences a force q 0 E . The displacement of the charge in moving from the initial position Pi to the final position Pf is the vector pointing from Pi to Pf , and is denoted by r . Δr Pf θ Pi E The work done by the electric force on the charge is W q0 E r cos where is the angle between the vector E and r . Defining the dot product between two vectors A and B as A B A B cos angle between A and B we can write W q0 E r . Note that in this formula q 0 can be either positive or negative. In the general case where the electric field is not uniform, and an arbitrary path is followed in moving the charge from Pi to Pf , the work done by the electric force along the path can be calculated by dividing the path into small segments of displacements, using the above formula with the local electric field for the work done in the small segment, and adding up the contributions: W q0 E r along the path from Pi 19 to Pj Pf Δr E Pi It can be shown that the electric force is conservative, which means that the work is independent of the path, and depends only on the two end points. Therefore, we can associate with each position of the test charge a quantity called its electric potential energy U in such a way that Ui U f W According to this definition, in going from a place with higher potential energy to one with lower potential energy U i U f , positive work is done by the electric field on the test charge. Conversely, in going from a point with lower to one with higher potential energy, the work done is negative. This is similar to the definition of gravitational potential energy U g mgh : in going from a higher position to a lower position on the earth’s surface, the gravitational attraction on a mass m does positive work. We now define the electric potential V at any point by V U q0 so that U q0V and observe that Vi V f E r along any path from Pi to Pf The right side of this equation can be regarded as the work done by the electric force on a unit positive test charge. Since only differences in potentials are involved in calculating work, the value of the electric potential is determined only after we have chosen certain point to be a reference point and call the electric potential zero there. Just as the electric field, the electric potential at a point does not depend on the test charge, and depends only on the point itself in relation to the charge distribution. The unit of electric potential is J/C(Joule/Coulomb), which is known as V (volt). 20 Specialized to the case of a uniform electric field, or applied to an infinitesimal path, we can write, in view of the customary meaning of the delta notation as final value minus the initial values, V E r which is a fundamental relation between the electric field and the electric potential. Electric potential in a region of uniform electric field A region of uniform electric field exists between the oppositely charged plates of a parallel plate capacitor as we have shown. Suppose the x-axis is chosen to be the direction of the electric field, whose x-component we denote by E x and is positive. We consider two points A,B so that the line AB is perpendicular to the x-axis. E A B Applying the fundamental relation, VA VB 0 because the displacement is perpendicular to the electric field. This shows that the potential on all points of a plane perpendicular to the electric field have the same electric potential. A surface on which all points have the same electric potential is called an equipotential surface. For a uniform electric field, we see that the equipotential surfaces are planes perpendicular to the field lines. Consider a point C “downwind” of the electric field from the point A and at a distance of x from it. E A C Then for the displacement from A to C, the increment of potential is V VC V A , so that according to the fundamental relation between potential and electric field, we have V A VC E x x 21 which is a positive number. Thus, locations “downwind” of the electric field have smaller electric potential. Also, electric field points in the direction from high to low electric potentials. The field lines (red) and equipotential surfaces (blue) inside a parallel plate capacitor are as shown: x d-x x d With the x-axis chosen as shown, and the electric potential chosen to be zero at the plate on the right, the potential V at a point whose coordinate is x satisfies the relation V Ed x where d is the distance between the plates. The graph of V against x is a straight line: V V0 x d The potential of the left plate is given by V0 Ed . Electric Potential of a Point Charge For a point charge q , zero potential is usually chosen to be at infinity. The potential V at a point P at a distance r from the charge can be calculated from the work done by the electric force on a unit positive test charge (or, by the electric field) for the displacement from P to a point at infinity. As we move from P to infinity, the electric field changes, and so the path has to be divided into many small parts and then summed. This process is known as integration. It leads to the result q V k r q r P V 0 22 Note that V has the same sign as q : a positive (negative) charge creates positive (negative) potential in its environment. Equipotential surfaces of a point charge are concentric spherical surfaces centered on the charge. The field lines, going out radially from the point charge, are perpendicular to the surfaces. The potential due to a distribution of point charges can be obtained from the sum of the potential due to the individual charges. Being a scalar, the potential is easier to calculate than the electric field. Example: For the dipole consisting of point charges of 5nC and 5nC 2m apart as shown, find the electric potentials at the points A and B. B θ 2m 5m m θ A -5nC 2m 1m 5nC m m Solution: Since the distances between A and positive and negative charges are 3m and 1m respectively, 5 10 9 15V Potential at A due to the positive charge is V L 9.0 10 9 3 5nC Potential at A due to the negative charge is VR 9.0 10 9 23 (5 10 9 ) 45V 1 Potential at A is V 15 45 30V The point B is at equal distance of V VL VR 9.0 10 9 5 10 9 5 5m from the two charges. The potential there is 9.0 10 9 (5 10 9 ) 5 0 Equipotential surfaces and field lines of a dipole can be found in the text book. Graphical Representation of Relation between Electric Potential and Electric Field In general, field lines are always perpendicular (normal) to equipotential surfaces. To prove this, we consider a point P on an equipotential surface and the electric field E at the point. We take another point Q on the same surface but very close to P, and let r denote the displacement from P to Q. Applying the fundamental relation V E r , we find E Q P E r 0 because Q and P have the same potential. This means the angle between the vectors E and r is 90º. Since the point Q can be in any direction away from P while still on the surface, we conclude that the direction of the field is normal to the equipotential surface In a 1-D problem, the electric potential depends only on the x coordinate and the electric field has only x-component. The fundamental relation V E r reduces to V E x x and can be rewritten as 24 Ex V x Thus in a plot of V against x , the negative of the slope at any point gives the xcomponent of the electric field at that point, as we have seen in the case of the parallel plate capacitor. This also holds true in the following two situations, although both are not 1-D problem. For a point charge q at the origin of the x-axis, the potential at any point on the x-axis is V x kq x Plots of V versus x and E x versus x are as shown: V Ex x x For a dipole made up of point charges q and q at a distance 2a apart, with the xaxis chosen so that the coordinates of the left and right charges are a and a respectively, the potential of an arbitrary point on the x-axis with coordinate x is xa -a a xa q q V x k k xa xa The plots of V and Ex versus x are as shown: Ex V x x 25 In a 2-D problem, the potential depends on both x and y coordinates. The equipotential surfaces are now curves in the x-y space. They are usually drawn so that neighboring curves differ by equal amount of potential. The electric field at any point P is perpendicular to the equipotential surface it is located at, and is in the direction pointing from the higher to the lower potential. The magnitude of the electric field is obtained from V E r where V is the potential difference between neighboring surfaces, and r is perpendicular distance between the neighboring surfaces at the point P. Decreasing V r P Thus, regions where the equipotential surfaces are closer together correspond to strong electric field. Electric Potential of Conductors. Since the electric field inside a conductor is zero in electrostatics, zero work is done by the electric field in going from one point to the other inside the conductor. Therefore, any two points in the conductor have the same electric potential. In fact, the whole conductor is an equipotential body. In particular, the surface of the conductor is an equipotential surface. The electric field line passing through any point on the surface is perpendicular to the surface. When multiple conductors are present, each is an equipotential body but the potentials are in general different. 26 5. Electric Potential II Potential Difference and Work Done on a Charge From the definition of electric potential, we deduce that when a charge q moves from an initial location where the electric potential is Vi to a final location where the potential is V f , the work done by the electric field is Wi f qVi V f Thus, in going from a place of high to one of lower electric potential, the electric field does positive work on a positive charge and negative work on a negative charge. When positive work is done on an object, its kinetic energy increases, otherwise the kinetic energy decreases, according to the work energy theorem. Example: A battery maintains a fixed potential difference between its two terminals because of the chemicals it contains. The potential is higher at the positive terminal. If a wire connects the terminals, the electric field inside the wire exerts a force on the electrons, causing them to move toward the positive terminal. Thus, negative charges are moved from a location of low potential (the negative terminal) to one of high potential (the positive terminal). The work done by the electric field on these charges is positive, which should cause increase of kinetic energy of the electrons. However, this increase does not take place when we have a steady flow of electrons: their speed does not change in going from one to the other terminals. This is because the electrons are not completely free. They collide with the positive ions, thereby experience a force of resistance. As a result, heat is evolved in the wire. The work done by the electric field is then equal to the heat energy evolved. Customarily, we replace the flow of negative charge by a flow of fictitious positive charge in the opposite direction, and the direction of electric current is taken to be the direction of motion of these fictitious positive charges. The work done by the electric field on these fictitious positive charges is equal to the real work done on the electrons. Suppose in a certain battery the potential difference between the terminals is kept at 12V, and the heat evolves at a rate of 60W in the wire connecting the terminals. How much charge has been transferred between the terminals in one hour’s of operation? Solution: Heat energy evolved = power ×time =60×3600=216,000 J Work done by electric field in moving a charge q = qVi V f q 12 27 Work done by electric field = heat evolved in circuit q 12 216,000 q 18,000C The energy unit electron volt (eV) Atomic particles carry charges in multiples of the electronic charge e 1.6 10 19 C . For their motion, it is convenient to introduce a unit of energy, known as the electron volt. The definition is 1eV 1.6 10 19 J Example: Find the work done in eV by the electric field on a helium nucleus moving from a point where the electric potential is 200V to another point where the potential is 600V. Solution: Using the fact that the charge of a helium nucleus is 2e , the work done by the electric field is W 2 1.6 10 19 200 600 J 2 200 600eV 800eV 2 1.6 10 19 200 600 / 1.6 10 19 eV Example: Find the speed of a 500eV electron. Solution: The kinetic energy of the electron is K 500eV 500 1.6 10 19 J 8.0 10 17 J From the relation 1 K me v 2 and me 9.1 10 31 kg for the mass of an electron, 2 the speed is v 2K me 2 8.0 10 17 1.3 10 7 m / s 31 9.1 10 Example What is the minimum work required to move an electron initially at rest at a distance of 5.29 10 11 m from the nucleus of a hydrogen atom to a point very far away from the nucleus? Solution Because the electron experiences an attractive force from the nucleus, an external force is required to move the electron away from the nucleus. If Wext denotes the work done by this force, from the work-energy theorem, the change of kinetic energy of the electron is 28 K Wext WE where WE is the work done by the electric force, and is given by WE eVi V f The potential at the initial location is e 1.6 10 19 9 Vi k 9.0 10 27.2V r 5.29 10 11 while at the final location which is very far away, V f 0 . Thus the work done by the electric force is WE eVi 1.6 10 19 27.2 J 27.2 eV and is negative. Since the initial kinetic energy K i is zero, the work energy theorem yields Wext K WE K f WE This work is minimum when K f 0 . It value is therefore Wext WE 27.2 eV In a hydrogen atom, the electron does not stay at rest at a distance r from the nucleus. Instead it travels in a circle around the nucleus with r as the radius. In this case, the initial kinetic energy is not zero. In fact, it can be shown that K i 13.6eV . Still with K f 0 , the minimum external energy to remove the electron from the atom is Wext K i WE 13.6 27.2 13.6 eV This is known as the ionisation energy of the hydrogen atom. Relation between electric force and potential We now consider the problem of determining the motion of a test charge q 0 in the vicinity of a charge distribution. We know that the force on the test charge is given by F q0 E , and E is in the direction of decreasing potential. Therefore, for a positive test charge, the electric force is in the direction of decreasing potential, while for a negative 29 test charge, the force is in the direction of increasing potential. In other words, positive charges seek lower potential and negative charges seek higher potential. We have also shown that the potential energy of the test charge is given by U q0V Therefore, for both positive and negative test charges, the force is in the direction of decreasing potential energy, as required by the concept of potential energy. Application of conservation of energy The law of conservation of energy for the motion of a charged particle in the electric field created by a charge distribution states that the sum of kinetic energy and electric potential does not change. It is a consequence of the work energy theorem and the conservative nature of the electric force. If the charge and mass of the particle are q and m respectively, the law can be written in the form 1 2 1 mv f qV f mvi2 qVi 2 2 where the subscripts on the symbol v for speed refer to the initial and final stages of the motion. An alternative form is 1 2 1 2 mv f mvi q Vi V f 2 2 which explicitly shows that only difference in potentials enters. Example A potential difference of 2kV is applied between the two plates of a parallel plate capacitor. A proton is introduced at rest on the positive plate. Find the speed of the proton when it reaches the negative plate. Solution We use energy conservation equation in the form K f K i qVi V f Since K i 0 , q e for proton, and Vi V f 2000V , we find K f 1.6 10 19 2000 3.2 10 16 J Using m 1.67 10 27 kg for the mass of a proton, the speed is 30 vf 2K f m 2 3.2 10 16 6.2 10 5 m / s 27 1.67 10 Example If an electron is introduced at the positive plate of the capacitor in the previous problem with a velocity of 10 7 m / s directed toward the negative plate, what fraction of the distance between the plates can the electron reach? Solution When the electron moves toward the negative plate, its kinetic energy is continuously reduced while the potential energy qV eV increases. But the electric potential V decreases. Let V be the drop in potential from the positive plate to the location where the electron stops. Applying energy conservation’ 0 K i eV Since Ki 1 1 me vi2 9.1 10 31 10 7 2 2 2 4.6 10 17 J V 2kV K 4.6 10 17 V i 288V e 1.6 10 19 x d Since the graph of potential against the distance from the positive plate is a straight line, the drop in potential is proportional to the distance from the positive plate. Using the fact that the total potential drop from the positive to the negative plate is 2000V, the fractional penetration of the electron is x V 0.14 d 2000 Example An alpha particle (another name for the helium nucleus, carrying a charge 2e ) with kinetic energy of K 5MeV travels directly toward a bare nucleus of an atom whose atomic number is Z (The charge of the nucleus is therefore Ze ). Find the distance of closest approach if the bare nucleus is gold. ( Z 79) Solution: The potential created by the bare nucleus at a distance r is V k Ze r 31 Let rmin denote the distance of closest approach. Applying energy conservation between this location ( K f 0 ) and the initial location which is infinitely far away Vi 0, K i K , Ze 0 2e k rmin rmin K 0 2kZe2 2 9.0 10 9 79 1.6 10 19 K 5 10 6 1.6 10 19 which is approaching the size of the nucleus. 32 2 4.6 10 14 m 6. Capacitance Definition of Capacitance A capacitor is a device to store electric charge. It also stores electric energy as we shall demonstrate. Conceptually, a capacitor is formed when we have two metallic conductors carrying equal and opposite variable charges Q and Q respectively. These charges create an electric field and electric potential in the environment, with the conductors forming equipotential bodies. We denote the potential difference between the conductors by V . This difference is defined as the potential of the positively charged conductor minus that of the negatively charged bodies, and its value is therefore positive. From Coulomb’s law, we expect V to be proportional to Q . The more the charge carried by the conductors, the more the potential difference between them. Therefore the quantity C Q V is independent of the charge Q carried by the conductors. It is known as the capacitance of the two conductors, and depends only on such geometrical factors as the sizes, shapes, and distances of the conductors. The unit of capacitance is F (farad), which is equal to coulomb per volt (C/V) Rearranging the above equation to read Q CV it follows that a large capacitance means large charge storage with modest potential difference. This is important as a large potential difference translates into a large electric field. When the electric field is too large, the medium surrounding the conductors can become conducting, and the capacitor discharges. The minimum electric field for discharge to happen is called the breakdown electric field. In air, the breakdown field is about one MV/m. Parallel Plate Capacitor This simplest capacitor is formed by two parallel conductor plates so close to each other that they can be considered as infinite plates as far as the calculation of the electric field between them is concerned. A formula for its capacitance can be obtained by calculating the potential difference V when the plates carry charges Q and Q . Because of attraction, these charges are concentrated on the inner sides of the metallic plates facing each other, forming sheet charges. If A denote the area of each side of the plate, the surface charge densities of the sheets are where 33 d Q A A A V The uniform electric field between the plates then has the magnitude E Q 0 0 A and points from the positive to the negative plates. Letting d be the distance between the plates, the potential difference is given by V Ed Substituting the equation for E in terms of Q , we find V Qd 0 A from which it follows from the definition of capacitance that C 0 A d and depends on geometrical factors only. Usually, the plates are charged by connecting them to the terminals of a dc power supply, such as a battery. The potential difference is fixed by the power supply. Example: Find the capacitance of a parallel plate capacitors made from circular plates of radius 10cm at a distance of 1mm apart in air. How much charge is stored when the plates are charged to a potential difference of 200V? How large is the electric field? Solution d 1mm 10 3 m 2 A r 2 0.1 0.0314m 2 34 C 0 A 0.0314 8.85 10 12 2.78 10 10 F 3 d 10 Q CV 2.78 10 10 200 5.56 10 8 C E V 200 3 2.00 10 5 V / m d 10 The energy of a capacitor To build up the charge on a capacitor to its final value Q starting from no charge, one can imagine a process of continually extracting positive charge from the negative plate and depositing it on the positive plate. Work is done by the external force during this process. The total work done in reaching the final configuration can be considered to be stored as electrical energy. To calculate the work done, suppose at one time during the build up, the capacitor carries the charge q , which is less than Q . At this time, the potential difference is denoted by v , which is less than the final potential difference V . When a small amount of charge q is moved from the negative to the positive plates, work done by external force Wext work done by electric force qv The total work ,which is equated to the energy stored, and is denoted by U is then U Wext vq q , the graph of v against q is a straight line as shown in the C figure. The summation above can be carried out by noticing that Wext is equal to the area underneath the graph. This area is that of a right angle triangle with sides equal to Q and V. v From the relation v V Therefore, U 1 QV 2 Δ q 35 q Q By using the relation Q CV , the energy can be written also in the forms U 1 CV 2 2 or U Q2 2C Example: How much energy is stored in the parallel plate capacitor in the previous example. Solution: U 1 1 CV 2 2.78 10 10 200 2 5.56 10 6 J 2 2 Energy stored in an electric field In the case of parallel plate capacitor, the stored energy can be rewritten as follows: 1 1 0 A Ed 2 1 0 E 2 Ad U CV 2 2 2 d 2 The factor Ad is equal to the volume of the space between the plates, which is only region with nonzero electric field. Therefore we can take the point of view that energy is stored in the space where there is an electric field, and that electric field energy density 1 0E2 2 Dielectrics A dielectric is a material made up of molecules that are permanent electric dipoles. When it is inserted between the plates of a parallel plate capacitor, the electric field causes the dipoles to be aligned, with the negative ends pointing to the positive plate. This results in a negative surface charge on the face of the dielectric that is in contact with the positive plate, and a positive surface charge on the face in contact with the negative plate. Inside 36 + + + + + + + + + - + - + - + - + - + - induced the dielectric, the charge is zero. We say that a surface charge is induced on the dielectric as a result of polarization induced by the electric field created by the charges on the plates. Let induced denote the surface charge density on the dielectric. Together with the charge on the plates, a pair of sheet charges with charge density induced are formed. This is reduced from the value without dielectric. Such an effect is known as screening. The ratio is a property of the dielectric called the dielectric constant. It is always greater than one. The electric field inside the capacitor is now due to these sheets charges. Its value is , 0 0 and is less than the value E0 0 without dielectric. E Q d d 0 0 A where we have used Q A . Thus, the capacitance is V Ed C Q A 37