3Electric Potential

Chapter 3 Electric Potential 3.1 Potential and Potential Energy.............................................................................. 3-2 3.2 Electric Potential in a Uniform Field.................................................................... 3-5 3.3 Electric Potential due to Point Charges ................................................................ 3-6 3.3.1 Potential Energy in a System of Charges....................................................... 3-8 3.4 Continuous Charge Distribution ........................................................................... 3-9 3.5 Deriving Electric Field from the Electric Potential ............................................ 3-10 3.5.1 Gradient and Equipotentials......................................................................... 3-11 Example 3.1: Uniformly Charged Rod ................................................................. 3-13 Example 3.2: Uniformly Charged Ring................................................................ 3-15 Example 3.3: Uniformly Charged Disk ................................................................ 3-16 Example 3.4: Calculating Electric Field from Electric Potential.......................... 3-18 3.6 Summary............................................................................................................. 3-18 3.7 Problem-Solving Strategy: Calculating Electric Potential.................................. 3-20 3.8 Solved Problems ................................................................................................. 3-22 3.8.1 Electric Potential Due to a System of Two Charges.................................... 3-22 3.8.2 Electric Dipole Potential .............................................................................. 3-23 3.8.3 Electric Potential of an Annulus .................................................................. 3-24 3.8.4 Charge Moving Near a Charged Wire ......................................................... 3-25 3.9 Conceptual Questions ......................................................................................... 3-26 3.10 Additional Problems ......................................................................................... 3-27 3.10.1 Cube ........................................................................................................... 3-27 3.10.2 Three Charges ............................................................................................ 3-27 3.10.3 Work Done on Charges.............................................................................. 3-27 3.10.4 Calculating E from V ................................................................................. 3-28 3.10.5 Electric Potential of a Rod ......................................................................... 3-28 3.10.6 Electric Potential........................................................................................ 3-29 3.10.7 Calculating Electric Field from the Electric Potential ............................... 3-29 3.10.8 Electric Potential and Electric Potential Energy........................................ 3-30 3.10.9. Electric Field, Potential and Energy .......................................................... 3-30 3-1 Electric Potential 3.1 Potential and Potential Energy In the introductory mechanics course, we have seen that gravitational force from the Earth on a particle of mass m located at a distance r from Earth’s center has an inverse- square form: 2 ˆg MmG r = −F rG (3.1.1) where is the gravitational constant and is a unit vector pointing radially outward. The Earth is assumed to be a uniform sphere of mass M. The corresponding gravitational field 11 2 26.67 10 N m /kgG −= × ⋅ rˆ gG , defined as the gravitational force per unit mass, is given by 2 ˆ g GM m r = = −Fg r G G (3.1.2) Notice that gG only depends on M, the mass which creates the field, and r, the distance from M. Figure 3.1.1 Consider moving a particle of mass m under the influence of gravity (Figure 3.1.1). The work done by gravity in moving from A to B is m 2 1 1 B B A A r r g g r B A r GMm r W d dr GMm r r GMm r − ⎛ ⎞⎛ ⎞= ⋅ = = = −⎜ ⎟ ⎜⎝ ⎠ ⎝ ⎠ ⎡ ⎤∫ ∫ ⎢ ⎥⎣ ⎦F s G G ⎟ (3.1.3) The result shows that gW is independent of the path taken; it depends only on the endpoints A and B. It is important to draw distinction between ,gW the work done by the 3-2 field and , the work done by an external agent such as you. They simply differ by a negative sign: . extW extgW W= − Near Earth’s surface, the gravitational field gG is approximately constant, with a magnitude , where is the radius of Earth. The work done by gravity in moving an object from height 2 2/ 9.8m/Eg GM r= ≈ s Er Ay to (Figure 3.1.2) is By cos cos ( )B A B B y g g B AA A y W d mg ds mg ds mg dy mg y yθ φ= ⋅ = = − = − = − −∫ ∫ ∫ ∫F sG G (3.1.4) Figure 3.1.2 Moving a mass m from A to B. The result again is independent of the path, and is only a function of the change in vertical height . B Ay y− In the examples above, if the path forms a closed loop, so that the object moves around and then returns to where it starts off, the net work done by the gravitational field would be zero, and we say that the gravitational force is conservative. More generally, a force F G is said to be conservative if its line integral around a closed loop vanishes: 0d⋅ =∫ F sG Gv (3.1.5) When dealing with a conservative force, it is often convenient to introduce the concept of potential energy U. The change in potential energy associated with a conservative force acting on an object as it moves from A to B is defined as: F JG B B A A U U U d W∆ = − = − ⋅ = −∫ F sG G (3.1.6) where W is the work done by the force on the object. In the case of gravity, gW W= and from Eq. (3.1.3), the potential energy can be written as 0g GMmU r U= − + (3.1.7) 3-3 where is an arbitrary constant which depends on a reference point. It is often convenient to choose a reference point where is equal to zero. In the gravitational case, we choose infinity to be the reference point, with 0U 0U 0 ( )U r 0= ∞ = . Since gU depends on the reference point chosen, it is only the potential energy difference gU∆ that has physical importance. Near Earth’s surface where the gravitational field gG is approximately constant, as an object moves from the ground to a height h, the change in potential energy is gU mgh∆ = + , and the work done by gravity is gW mgh= − . A concept which is closely related to potential energy is “potential.” From , the gravitational potential can be obtained as U∆ ( / ) Bg g gA U V m d m ∆∆ = = − ⋅ = − ⋅∫ F s BA d∫ g sG G G G (3.1.8) Physically gV∆ represents the negative of the work done per unit mass by gravity to move a particle from . to A B Our treatment of electrostatics is remarkably similar to gravitation. The electrostatic force given by Coulomb’s law also has an inverse-square form. In addition, it is also conservative. In the presence of an electric field E eF JG JG , in analogy to the gravitational field gG , we define the electric potential difference between two points as and A B 0( / ) B eA V q d∆ = − ⋅ = − ⋅∫ ∫F s EBA d sG GG G (3.1.9) where is a test charge. The potential difference 0q V∆ represents the amount of work done per unit charge to move a test charge from point A to B, without changing its kinetic energy. Again, electric potential should not be confused with electric potential energy. The two quantities are related by 0q 0U q V∆ = ∆ (3.1.10) The SI unit of electric potential is volt (V): (3.1.11) 1volt 1 joule/coulomb (1 V= 1 J/C)= When dealing with systems at the atomic or molecular scale, a joule (J) often turns out to be too large as an energy unit. A more useful scale is electron volt (eV), which is defined as the energy an electron acquires (or loses) when moving through a potential difference of one volt: 3-4 (3.1.12) 19 191eV (1.6 10 C)(1V) 1.6 10 J−= × = × − 3.2 Electric Potential in a Uniform Field Consider a charge q+ moving in the direction of a uniform electric field 0 ˆ( )E= −E j JG , as shown in Figure 3.2.1(a). (a) (b) Figure 3.2.1 (a) A charge q which moves in the direction of a constant electric field E JG . (b) A mass m that moves in the direction of a constant gravitational field gG . Since the path taken is parallel to E JG , the potential difference between points A and B is given by 0 0 0 B B B A A A V V V d E ds E d∆ = − = − ⋅ = − = − <∫ ∫E sJG G (3.2.1) implying that point B is at a lower potential compared to A. In fact, electric field lines always point from higher potential to lower. The change in potential energy is . Since we have0B AU U U qE d∆ = − = − 0,q > 0U∆ < , which implies that the potential energy of a positive charge decreases as it moves along the direction of the electric field. The corresponding gravitational analogy, depicted in Figure 3.2.1(b), is that a mass m loses potential energy ( ) as it moves in the direction of the gravitational field U mg∆ = − d gG . Figure 3.2.2 Potential difference due to a uniform electric field What happens if the path from A to B is not parallel to E JG , but instead at an angle θ, as shown in Figure 3.2.2? In that case, the potential difference becomes 3-5 0 cos B B A A V V V d E s E yθ∆ = − = − ⋅ = − ⋅ − = −∫ E s E s = 0JG JGG G (3.2.2) Note that y increase downward in F gure 3.2.2. Here we see once more that moving along the direction of the electric field E iJG leads to a lower electric potential. What would the change in potential be if the path were ? In this case, the potential difference consists of two contributions, one for each segment of the path: A C B→ → CA BCV V V∆ = ∆ + ∆ (3.2.3) When moving from A to C, the change in potential is 0CAV E y∆ = − . On the other hand, when going from C to B, since the path is perpendicular to the direction of E0BCV∆ = JG . Thus, the same result is obtained irrespective of the path taken, consistent with the fact that E JG is conservative. Notice that for the path , work is done by the field only along the segment AC which is parallel to the field lines. Points B and C are at the same electric potential, i.e., . Since , this means that no work is required in moving a charge from B to C. In fact, all points along the straight line connecting B and C are on the same “equipotential line.” A more complete discussion of equipotential will be given in Section 3.5. A C B→ → BV V= C r U q V∆ = ∆ 3.3 Electric Potential due to Point Charges Next, let’s compute the potential difference between two points A and B due to a charge +Q. The electric field produced by Q is 20 ˆ( / 4 )Q rπε=E JG , where is a unit vector pointing toward the field point. rˆ Figure 3.3.1 Potential difference between two points due to a point charge Q. From Figure 3.3.1, we see that ˆ cosd ds drθ⋅ = =r sG , which gives 2 2 0 0 0 1 1ˆ 4 4 4 B B B A A A B A Q Q QV V V d dr r rπε πε πε ⎛ ⎞∆ = − = − ⋅ − = −⎜⎝ ⎠∫ ∫r s = G r r ⎟ (3.3.1) 3-6 Once again, the potential difference V∆ depends only on the endpoints, independent of the choice of path taken. As in the case of gravity, only the difference in electrical potential is physically meaningful, and one may choose a reference point and set the potential there to be zero. In practice, it is often convenient to choose the reference point to be at infinity, so that the electric potential at a point P becomes P PV ∞ d= − ⋅∫ E sJG G (3.3.2) With this reference, the electric potential at a distance r away from a point charge Q becomes 0 1( ) 4 QV r rπε= (3.3.3) When more than one point charge is present, by applying the superposition principle, the total electric potential is simply the sum of potentials due to individual charges: 0 1( ) 4 i i e i ii i q qV r k r rπε= =∑ ∑ (3.3.4) A summary of comparison between gravitation and electrostatics is tabulated below: Gravitation Electrostatics Mass m Charge q Gravitational force 2 ˆg MmG r = −F rG Coulomb force 2 ˆe e Qqk r=F r G Gravitational field /g m=g F GG Electric field /e q=E F G G Potential energy change B gA U∆ = − ⋅∫ F sdG G Potential energy change B eAU d∆ = − ⋅∫ F sG G Gravitational potential B g A V d= − ⋅∫ g sG G Electric Potential BAV d= − ⋅∫ E sG G For a source M: g GMV r = − For a source Q: e QV k r= | |gU mg∆ = d (constant gG ) | |U qEd∆ = (constant E ) JG 3-7 3.3.1 Potential Energy in a System of Charges If a system of charges is assembled by an external agent, then extU W W∆ = − = + . That is, the change in potential energy of the system is the work that must be put in by an external agent to assemble the configuration. A simple example is lifting a mass m through a height h. The work done by an external agent  you, is mgh+ (The gravitational field does work ). The charges are brought in from infinity without acceleration i.e. they are at rest at the end of the process. Let’s start with just two charges and . Let the potential due to at a point be (Figure 3.3.2). mgh− 1q 2q 1q P 1V Figure 3.3.2 Two point charges separated by a distance . 12r The work done by an agent in bringing the second charge from infinity to P is then . (No work is required to set up the first charge and ). Since 2W 2q 2 2W q V= 1 1 0W = 1 1 0 12/ 4 ,V q rπε= where is the distance measured from to P, we have 12r 1q 1 212 2 0 12 1 4 q qU W rπε= = (3.3.5) If and q1q 2 have the same sign, positive work must be done to overcome the electrostatic repulsion and the potential energy of the system is positive, . On the other hand, if the signs are opposite, then due to the attractive force between the charges. 12 0U > 12 0U < Figure 3.3.3 A system of three point charges. To add a third charge q3 to the system (Figure 3.3.3), the work required is 3-8 ( ) 3 1 23 3 1 2 0 13 234 q q qW q V V r rπε ⎛ ⎞= + = +⎜⎝ ⎠⎟ (3.3.6) The potential energy of this configuration is then 1 3 2 31 22 3 12 13 23 0 12 13 23 1 4 q q q qq qU W W U U U r r rπε ⎛ ⎞= + = + + = + +⎜ ⎟⎝ ⎠ (3.3.7) The equation shows that the total potential energy is simply the sum of the contributions from distinct pairs. Generalizing to a system of N charges, we have 0 1 1 1 4 N N i j iji j j i q q U rπε = => = ∑∑ (3.3.8) where the constraint j i> is placed to avoid double counting each pair. Alternatively, one may count each pair twice and divide the result by 2. This leads to 0 01 1 1 1 1 1 1 1 1 ( ) 8 2 4 2 N N N N N i j j i ij iji j i j i j i j i q q q U q r rπε πε= = = = =≠ ≠ ⎛ ⎞⎜ ⎟= = =⎜ ⎟⎜ ⎟⎝ ⎠ ∑∑ ∑ ∑ ∑ i iqV r (3.3.9) where , the quantity in the parenthesis, is the potential at ( )iV r ir G (location of qi) due to all the other charges. 3.4 Continuous Charge Distribution If the charge distribution is continuous, the potential at a point P can be found by summing over the contributions from individual differential elements of charge . dq Figure 3.4.1 Continuous charge distribution 3-9 Consider the charge distribution shown in Figure 3.4.1. Taking infinity as our reference point with zero potential, the electric potential at P due to dq is 0 1 4 dqdV rπε= (3.4.1) Summing over contributions from all differential elements, we have 0 1 4 dqV rπε= ∫ (3.4.2) 3.5 Deriving Electric Field from the Electric Potential In Eq. (3.1.9) we established the relation between E JG and V. If we consider two points which are separated by a small distance dsG , the following differential form is obtained: dV d= − ⋅E sJG G (3.5.1) In Cartesian coordinates, ˆ ˆ ˆx y zE E E= + +E i j k JG and ˆ ˆ ˆ ,d dx dy dz= + +s i j kG we have ( ) ( )ˆ ˆ ˆ ˆˆ ˆx y z x y zdV E E E dx dy dz E dx E dy E dz= + + ⋅ + + = + +i j k i j k (3.5.2) which implies , ,x y z V VE E E V x y z ∂ ∂= − = − = − ∂∂ ∂ ∂ (3.5.3) By introducing a differential quantity called the “del (gradient) operator” ˆ ˆ ˆ x y z ∂ ∂ ∂∇ ≡ +∂ ∂ ∂i j+ k (3.5.4) the electric field can be written as ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ x y z V V VE E E V V x y z x y z ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂= + + = − + = − + = −∇⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠E i j k i + j k i + j k JG V= −∇EJG (3.5.5) Notice that ∇ operates on a scalar quantity (electric potential) and results in a vector quantity (electric field). Mathematically, we can think of E JG as the negative of the gradient of the electric potential V . Physically, the negative sign implies that if 3-10 V increases as a positive charge moves along some direction, say x, with , then there is a non-vanishing component of E / 0V x∂ ∂ >JG in the opposite direction ( . In the case of gravity, if the gravitational potential increases when a mass is lifted a distance h, the gravitational force must be downward. 0)xE− ≠ If the charge distribution possesses spherical symmetry, then the resulting electric field is a function of the radial distance r, i.e., ˆrE=E r G . In this case, .rdV E dr= − If is known, then E may be obtained as ( )V rG ˆr dVE dr ⎛ ⎞= = −⎜ ⎟⎝ ⎠E r rˆ JG (3.5.6) For example, the electric potential due to a point charge q is 0( ) / 4V r q rπε= . Using the above formula, the electric field is simply 20 ˆ( 4 )q rπε=E / r JG . 3.5.1 Gradient and Equipotentials Suppose a system in two dimensions has an electric potential . The curves characterized by constant are called equipotential curves. Examples of equipotential curves are depicted in Figure 3.5.1 below. ( , )V x y ( , )V x y Figure 3.5.1 Equipotential curves In three dimensions we have equipotential surfaces and they are described by =constant. Since we can show that the direction of E JG is always perpendicular to the equipotential through the point. Below we give a proof in two dimensions. Generalization to three dimensions is straightforward. ( , , )V x y z ,V= −∇EG Proof: Referring to Figure 3.5.2, let the potential at a point be . How much is changed at a neighboring point ( , )P x y ( , )V x y V ( , )P x dx y dy+ + ? Let the difference be written as 3-11 ( , ) ( , ) ( , ) ( , ) dV V x dx y dy V x y V V V VV x y dx dy V x y dx dy x y x y = + + − ⎡ ⎤∂ ∂ ∂ ∂= + + + − ≈ +⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦" (3.5.7) Figure 3.5.2 Change in V when moving from one equipotential curve to another With the displacement vector given by ˆd dx dy= +s i jˆG , we can rewrite as dV ( )ˆ ˆ ˆ ˆ ( )V VdV dx dy V d dx y⎛ ⎞∂ ∂= + ⋅ + = ∇ ⋅ = − ⋅⎜ ⎟∂ ∂⎝ ⎠i j i j s E s JG G (3.5.8) If the displacement d is along the tangent to the equipotential curve through P(x,y), then because V is constant everywhere on the curve. This implies that sG 0dV = d⊥E sJG G along the equipotential curve. That is, E JG is perpendicular to the equipotential. In Figure 3.5.3 we illustrate some examples of equipotential curves. In three dimensions they become equipotential surfaces. From Eq. (3.5.8), we also see that the change in potential attains a maximum when the gradientdV V∇ is parallel to d sG : max dV V ds ⎛ ⎞ = ∇⎜ ⎟⎝ ⎠ (3.5.9) Physically, this means that always points in the direction of maximum rate of change of V with respect to the displacement s. V∇ Figure 3.5.3 Equipotential curves and electric field lines for (a) a constant E field, (b) a point charge, and (c) an electric dipole. JG