Electric Fields Part 1: Coulomb s Law

Transcription

1 Electric Fields Part 1: Coulomb s Law F F Last modified: 07/02/2018

2 Contents Links Electric Charge & Coulomb s Law Electric Charge Coulomb s Law Example 1: Coulomb s Law Electric Field Electric Field Vector Electric Field of a Point Charge Example 2: Electric Field of Single Charge Electric Field of Multiple Charges Example 3: Field due to Multiple Charges Why Bother With Field Vectors? Gravitational Field Electric Field Lines Electric Potential Definition Example 4: Charge & Potential Difference Potential due to a Point Charge Example 5: Potential due to Multiple Charges Equipotentials Summary: Electric Field & Potential

3 Electric Charge Contents Electric charge is a basic property of matter. Protons (positive) and electrons (negative) have equal and opposite signed charges. An object will normally have equal numbers of protons and electrons and so have zero total charge ( neutrally charged or just neutral ), but by adding/removing electrons from an object (perhaps by rubbing it against another object) we can give it a net negative/positive electric charge. The SI unit of charge is the coulomb (C) The charge of a proton is traditionally represented by the letter e, where e = C (and of course the electron charge is e = C). Because e is so very small, real world charges often need to be measured in mc, µc, nc or even pc and fc. (i.e. 10 3, 10 6, 10 9, 10 12, C)

4 Coulomb s Law Contents Two different charges a distance apart will both experience a force. If the charges are of the same sign (i.e. both positive or both negative) then the forces will be repulsive - pushing the charges apart. F F and F F If the charges have opposite signs, then the forces will be attractive - pulling the charges together. F F The forces on the two charges are equal in magnitude, but opposite in direction (as we should expect from Newton s third law).

5 Contents In 1785 Charles Coulomb performed a series of careful experiments, to determine a formula for the magnitude of the forces acting on two charges Q 1 and Q 2 separated by a distance r. (Traditionally charges are represented by the letter Q or q) F Q 1 r Q 2 F These experiments led to Coulomb s Law for the magnitude of the force between the charges: r Q 1 F Q 2 F F = k Q 1Q 2 r 2 k = Nm 2 C 2 is known as Coulomb s constant. Note that changing the sign of a charge doesn t change the size of the forces, only possibly their direction.

6 Example 1 Contents Two charges of 5 µc and 2 µc respectively are located 2 m apart. Calculate the magnitude of the force acting on the negative charge. The two opposite charges will be attracted by equal and opposite forces. F 5 µc 2 µc r = 2 m F The magnitude F of both forces is given by Coulomb s Law: F = k Q 1Q 2 r 2 = ( ) ( )( ) 2 2 = N

7 Electric Field Vector Consider a stationary positive charge: A second positive charge q placed a distance r away will experience a force F, given by Coulomb s Law: A different positive charge q placed at the same point will experience a force F, again given by Coulomb s Law: And a negative charge q at this same point will experience a force F, towards the charge Q: Q Contents F = q ( ) kq r Q q 2 r F = q ( ) kq Q q r 2 r F = q ( ) kq r 2 Q q r Note that in each case, though the sizes of the forces vary, the term in brackets, kq r 2, is the same.

8 Contents Now, let s introduce a new vector E with magnitude equal to this term, and directed away from the charge: Q r E = kq r 2 For each of the previous examples, we can now express the force on the charge in terms of this new vector E: F = qe Q q E F = q E Q q Q E F = q E q E Repeating these steps for a negative charge, we find the only difference is that the vector E changes direction: F = qe Q q E F = q E Q q Q E F = q E E q

9 Electric Field of a Point Charge Contents The vector E is called the electric field vector. We can imagine this electric field vector being present at all points around a point charge (i.e. a very physically small charge). P Q r Positive charge P Q r Negative charge In each case, the magnitude of the field vector at a point P is given by the same expression: E(r) = k Q r The force acting on a charge q at point P will be F = qe The SI unit of electric field is newton/coulomb (N/C).

10 Example 2 Contents A point charge of Q = 5 µc is located at the origin (0,0). Determine the electric field vector E at the point P = (1.2, 1.6), in both polar and Cartesian forms. Hence calculate the magnitude of the Coulomb force acting on a charge q = 2 µc when it is placed at P. r = 2 m 5 µc θ 1.2 m P E 1.6 m E will be directed away from the positve charge and its magnitude will be: E = k Q r 2 = ( ) = N/C and θ = tan to x-axis.

11 Contents In Cartesian form, the field vector is: E = 11250(cos 53 i sin 53 j) = 11250( 3 5 i 4 5 j) = 6750 i 9000 j N/C When placed at the point P, charge q experiences a force F as shown at right. The magnitude of this force F is: F = q E = ( )11250 = N 5 µc r = 2 m -2 µc θ 1.2 m F E 1.6 m As we should have expected, this is the same result we found for the same charges in Example 1.

12 Electric Field of Multiple Charges Contents Consider a region containing two point charges. Q 2 At a point P there will be two electric field vectors as shown - one produced by each charge. Q 1 P E 2 E 1 If a charge q is now placed at P, then it will experience two forces. The resultant force on charge q is: Q 2 F resultant = F 1 F 2 = qe 1 qe 2 = q(e 1 E 2 ) Q 1 = q E resultant F 1 = qe 1 q F 1 F 2 F 2 = qe 2 The total electric field vector at P is the sum of the point charge field vectors. This will of course also be true for 3, 4, 5... charges.

13 Example 3 Contents A point charge of 5 µc is located at the origin (0,0). A second charge of 3 µc is at the point (0,1.6). (a) Determine the total electric field vector E P at the point P = (1.2, 1.6). (b) If a third charge of 2 µc is placed at the point P, what Coulomb force would it feel? The total field E P will be the sum of the individual fields from each of the two charges: E P = E E E P 3 µc E 2 m 5 µc 1.2 m P 1.6 m E E has already been calculated (in the previous example), so next we must determine the second field vector E. E = k Q r 2 = N/C = (9 109 ) (1.2) 2

14 Contents Direction is the negative x-axis, so E = i N/C The total field E P is then calculated to be: E P = E E = (6750 i 9000 j) ( i) N/C = i 9000 j N/C Or in polar form (check this for yourself!): N/C at nc E P P F = qep The force on a charge q placed at P will be F = qe P, so for q = 2 µc : F = ( 2.4 i 1.8 j) 10 2 N 5 nc or: N at 143

15 Why Bother With Field Vectors? Can t We Just use Coulomb s Law? Yes we could, but... the electric field concept helps us to answer an important question about Coulomb forces. Two charges will experience forces, even though they are separated. How does each charge know about the other? Contents We can answer this question in terms of the electric field as follows: A charge Q creates an electric field vector E at every point around it. These vectors sit, doing nothing, until... a second charge q is placed at a point in the field, where it feels a force F = qe Q q F This will also work in reverse - the charge q creates a field which causes a force on Q. We have actually already been using exactly this idea, but without using the word field.

16 Gravitational Field Contents As we know, the Earth s gravity can act on a mass even when the mass is not in contact with the ground. We can ask the same question - how does this mass know about the Earth? - and the answer is very similar: The Earth (or actually any mass) will create a gravitational field vector g at every point around it. These vectors just exist, doing nothing, until... a mass m is placed at some point in the field, where it will feel a force F = mg m F g Note the pattern, Gravity: F = m g property that feels force field vector Coulomb: F = q E

17 Electric Field Lines We can represent the direction of the electric field around a charge with a so-called quiver plot as seen at right. The direction of the field at different points is represented by little arrows at those points. Note that the field is of course three dimensional - the diagram showing the field only in the plane of the page. The direction of the field vector is always in the same direction as the Coulomb force that would act on a positive charge at that point. It can be a bit neater to instead indicate the general shape of the electric field using electric field lines. For the single charge, at right, these field lines are straight, but in general could be curved. The field vector at a particular point is always tangent to the field line at that point, such as at point P in the diagram. Contents Field Vectors P E Field Lines

18 Contents We ve looked here at the field around a positive charge, but the same logic applies to a negative charge. All that changes is the direction of the field vector. Field Vectors Field Lines For a single charge the magnitude of the field increases closer to the charge. This is indicated on the diagram by the field lines being closer together. Note also that field lines begin on a positive charge, and terminate on a negative charge. If there is a net charge, then the field lines will instead start/finish at infinity.

19 Contents The electric field lines around a collection of multiple charges can be determined by adding together the fields of each point charge. Two relatively simple examples are shown below. At left is the field of a electric dipole - 2 equal and opposite separated charges, and at right, the field of two equal charges. Remember again, that these diagrams show a 2D slice of a 3D field. field of electric dipole field of 2 equal charges

20 Electric Potential Contents Consider a small charge q located a distance r from a larger, fixed charge Q. Work must be done to place the charge at this point. This work will be the potential energy of the charge. Q r q F = qe If the charge q is free to move, then it will be repelled by the Coulomb force F and thus accelerate away from Q. The Coulomb force is doing work on the charge, converting potential energy to kinetic energy. (This situation is similar to a mass being dropped from a height above the ground, where gravity converts PE to KE). If we can calculate the gain in kinetic energy of the small charge (i.e. the work done), this will be equal to the loss in potential energy of the charge.

21 Contents Calculation of the work done is more complicated than in the gravitational case because the magnitude of the force F is varying with distance. However over a small displacement x we can take the force to be constant. The work done is: W = F x = F x = qe x q x F This is a loss of potential energy: Dividing by q: PE = qe x V = PE q = E x where V is called the potential difference between the start and end of the displacement. The SI unit of potential difference is the volt (V) 1 V = 1 J/C

22 Contents Work W = q V is done on a charge q moving through a potential difference V. If V is positive, this means an increase in potential energy. If V is negative, kinetic energy will increase. From the previous page: taking the limit x 0 : V = E x E = V x E = dv dx The SI unit for electric field is also the volt per metre (V/m) V/m and N/C are both valid SI units for electric field, but V/m is the more commonly used. The above formula only applies for one dimension. In three dimensions it becomes: E = V x i V y j V z k

23 Example 4 Contents Two points A and B are located a distance 2 m 2 m apart. The potential difference V between A and B is 100 V. A B (a) A charge Q 1 = 5 µc moves from A to B. How much work is done on this charge? (b) A charge Q 2 = 4 µc moves from A to B. How much work is done on this charge? (c) If the electric field between the points is uniform, then what is the magnitude of this field? Its direction? (d) If the charge Q 2 is placed at the point midway between A and B, what is the magnitude and direction of the Coulomb force on this charge?

24 Contents (a) The work W done on the charge is: W = Q 1 V = ( )(100) = J This work represents a gain in potential energy. This must be either work done by an external force, or a corresponding loss in kinetic energy. (This is similar to the situation of a mass m, increasing its height.) (b) The work W done on the charge is: W = Q 2 V = ( )(100) = J This work represents a loss in potential energy. There will be a corresponding gain in kinetic energy. (This is similar to the situation of a mass m, falling downward.) Like gravity, the Coulonb force is conservative.

25 Contents (c) Because the electric field is uniform (i.e. constant) between A and B, then it is given by: E = V x = 100 = 50 V/m A B 2 The electric field is 50 V/m, directed from B to A. (d) The force on a charge anywhere between A to B will be the same: F = Q 2 E = ( )( 50i) = N A E F = Q 2 E B As we should have expected, the force on the charge is in the opposite direction to the field, i.e. towards B. Q 2

26 Potential due to a Point Charge Contents If a charge q is placed at a point P, located at a distance r from a fixed charge Q, it will be repelled and in the absence of other forces, will continue moving until it is an infinite distance away from Q. Consider the moment when q is a distance x away from the fixed charge Q: Q r P x x q F The work W done by the force F in moving the charge q through an additional small displacement x will be: W = F x = F x = k qq x 2 x

27 Contents To calculate the total work done in travelling from P (where x = r) to x = we need to add the small amounts of work one for each small section of this displacement and take the limit as x 0: W = lim ( W = kqq lim x 0 x=r ] = kqq [ 1 x r x 0 x=r [ = kqq 1 1 r ] x x 2 ) = kqq r dx = kqq r x 2 The amount of work done is a loss of potential energy and a gain in kinetic energy. (Compare this to an object falling due to gravity) So the change in potential energy is PE = kqq r, and recalling the definition of potential difference: V = PE q = kq r

28 Contents This potential difference is the difference between the potential at the final position V and the starting position V P : V = V V P = kq r As you should remember from last Semester, we can choose to define zero potential energy to be wherever we find convenient. In the case of electrical potential energy (and thus potential), the most convenient point is at r =, so V = 0 and we have: V P = kq r This expression will apply to any point around the charge Q. The work required to place a charge q at P (from infinitely far away) is: qv P = kqq r This is of course also the potential energy of the charge q at this point.

29 Contents The expression for the potential V at a point located a distance r from a charge Q is: V (r) = kq r V (r) does not have direction. It forms a scalar field around the charge Q. It s official name is scalar electromagnetic potential. Note that if Q is negative, then the potential will also be negative. (The negative sign indicates that we would need to add energy to move a positive charge to infinity from a point r from the charge) If more than one charge is present, the total potential will be simply the sum of the individual potentials calculated with the above formula. (This is because work/energy adds together).

30 Example 5 Contents A point charge of 5 µc is located at the origin (0,0). A second charge of 3 µc is at the point (0,1.6). (a) Calculate the total electric potential V at the point P = (1.2,1.6). (b) How much work must be done to place a third charge of 2 µc at the point P? The total potential V will be the sum of the potentials from each of the two charges. 3 µc 1.2 m 2 m P For the positive charge: V = kq r = (9 109 )( ) 2 5 µc = V

31 Contents and for the negative: V = kq r = (9 109 )( ) 1.2 Thus the total potential at P will be: = V V P = V V = ( 22500) = 0 V The work W required to place a charge at P will be W = qv P = 0 J for any charge q!

32 Equipotentials Contents Because the potential depends only on r, the equipotential surfaces (i.e. collection of points with the same potential) around a point charge will be surfaces with equal r, i.e. spheres. Q equipotential surfaces field lines and the equipotentials are perpendicular to each other closer spacing of the field lines indicates a larger field strength. closer spacing of the equipotentials indicates a larger field strength. field lines flow from high potential to low potential.

33 Contents If we know the electric field configuration of some charges, then we can draw in equipotentials, knowing they must be perpendicular to the electric field lines. For example, the equipotentials for the two examples seen previously: electric dipole 2 equal charges

34 Summary: Electric Field & Potential Contents The electric field lines around a point charge are directed outwards for a positive charge and inwards for a negative. E P Q P E Q The electric field vector E at a point P, located a distance r from the charge is tangent to the field lines, with magnitude: E(r) = k Q r 2 (SI unit: V/m or N/C) The Coulomb force acting on a charge q placed at this point P is given by: F = qe

35 Contents The scalar electric potential V (r) at the same point is: V (r) = k Q r (SI unit: V) V (r) is positive when Q is positive, and negative when Q is negative. The work, W done to place a charge q at the point P (i.e. the potential energy of the charge at that point) is given by: W = PE = qv = k qq r

36 Contents In a situation with multiple charges: The total electric field vector at a point is the vector sum of the individual vectors from each charge. The total potential at a point is the sum of the individual potentials from each charge. If a charge q is placed at a point with field vector E and scalar potential V, then: The Coulomb force on the charge will be F = qe. The work done to place the charge (i.e. the potential energy of the charge) will be qv.

37 Contents Properties of Field Lines and Equipotentials For any field, not just electric, the following properties are always true: Field lines are continuous. Field lines do not intersect each other. Field lines and Equipotentials are perpendicular to each other. Field lines are always directed from high potential to low potential. Field lines or Equipotentials being closer together indicates a stronger Field.