PH 222-2A Spring 2015

PH 222-2A Spring 215 Electric Potential Lectures 5-6 Chapter 24 (Halliday/Resnick/Walker, Fundamentals of Physics 9 th edition) 1

Chapter 24 Electric Potential In this chapter we will define the electric potential ( symbol V ) associated with the electric force and accomplish the following tasks: Calculate V if we know the corresponding electric field. Calculate the electric field if we know the corresponding potential V. Determine the potential V generated by a point charge. Determine the potential V generated by a discrete charge distribution. Determine the potential V generated by a continuous charge distribution. Determine the electric potential energy U of a system of charges. Define the notion of an equipotential surface. Explore the geometric relationship between equipotential surfaces and electric field lines. Explore the potential of a charged isolated conductor. 2

... O A U x f F( x) dx x i F(x) x i x x f f i U q Eds x B Electric Potential Energy In Chapter 8 we defined the change in potential energy U associated with a conservative force as the negative value of the work W that the force must do on a particle to take it from an initial position x to a final position x. i U U f Ui W F( x) dx Consider an electric charge q moving from an initial position at point A to a final position at point B under the influence of a known electric field E. The force exerted on the charge is F q E. f f U F ds q E ds i x x f i i f 3

A VP P E ds The Electric Potential V The change in potential energy of a charge q moving under the influence of E from point A B f to point B is: U U U W q Eds. f Please note that U depends on the value of q. i i We define the electric potential V in such a manner so that it is independent f U W of q: V Here V Vf Vi Vf Vi Eds. q q In all physical problems only changes in V are involved. Thus we can define arbitrarily the value of V at a reference point, which we choose to be at infinity: V V. We take the initial position as the generic point P with potential V : V f P P E ds. The potential V depends only on the coordinates of P and on E. P i P 4

V P 1 4 q R SI Units of V : Definition of voltage : Units of V: J/C, known as the volt Potential Due to a Point Charge V W q Consider a point charge q placed at the origin. We will use the definition given on the previous page to determine the potential V at point P a distance R from O. P R V E ds Edr cos Edr P R The electric field generated by q is: R E q 4 r 2 O V P q 4 R dr 2 r dx 2 x 1 x q 1 1 VP 4 r 4 R q R 5

Example. The potential of a point charge for a zero reference potential at infinity A point charge of q=4.x1-8 C creates a potential at a spot 1.2 m away. The potential is (a) V=kq/r= =(8.99x1 9 )(+4.x1-8 )/1.2=+3V when the charge is positive and (b) -3V when the charge is negative 6

The acceleration of positive and negative charges A positive charge accelerates from a region of higher electric potential toward a region of lower electric potential A negative charge accelerates from a region of lower electric potential toward a region of higher electric potential 7

Electrostatic Potential of a point charge 8

The electric potential energy of a point charge q for non-uniform electric field 9

One electron volt is the magnitude of the amount by which the potential energy of an electron changes when the electron moves through the potential difference of 1 volt 1

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A headlight connected to a 12-V battery 12

-2.3x1-18 J -6.5x1-19 J done by the electric force Work to be done against the electric force = +1 ev 13

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q 2 q 3 q 1 r 2 r 3 r 1 V V1V2 V3 P Potential Due to a Group of Point Charges Consider the group of three point charges shown in the figure. The potential V generated by this group at any point P is calculated using the principle of superposition. 1. We determine the potentials V, V, and V generated by each charge at point P: 1 2 3 1 q 1 q 1 V, V, V 1 2 3 1 2 3 4 r1 4 r2 4 r3 2. We add the three terms: V V V V V 1 2 3 1 q 1 q 1 4 r 4 r 4 1 2 1 2 q 3 r 3 q The previous equation can be generalized for n charges as V 1 q 1 q 1 q 1 q r r r n 1 2 n... 4 1 4 2 4 rn 4 1 i i follows: 16

Example : Potential Due to an Electric Dipole Consider the electric dipole shown in the figure. p We will determine the electric potential V created at point by the two charges of the dipole using superposition. Point P is at a distance r from the center O of the dipole. Line OP makes an angle with the dipole axis: P A 1 q q q r( ) r( ) V V( ) V( ). 4 r( ) r ( ) 4 r( ) r( ) We assume that r d, where d is the charge separation. From triangle ABC we have: r r d cos. ( ) ( ) q dcos 1 pcos Also: r r r V, 2 ( ) ( ) 2 2 4 r 4 r where p qd the electric dipole moment. B C V 1 p cos 2 4 r 17

r. P Potential Due to a Continuous Charge Distribution Consider the charge distribution shown in the figure. In order to determine the electric potential V created dq by the distribution at point P we use the principle of superposition as follows: 1. We divide the distribution into elements of charge dq. For a volume charge distribution, dq dv. V 1 4 dq r For a surface charge distribution, dq da. For a linear charge distribution, dq d. 1 dq 2. We determine the potential dv created by dq at P: dv. 4 r 1 dq 3. We sum all the contributions in the form of the integral: V. 4 r Note 1 : The integral is taken over the whole charge distribution. Note 2 : The integral involves only scalar quantities. 18

Example : Potential created by a line of charge of length L and uniform linear charge density λ at point P. Consider the charge element dq dx at point A, a distance x from O. From triangle OAP we have: 2 2 r d x d OP. Here is the distance. The potential dv created by dq at P is: dv 1 dq 1 dx 4 r 4 d x 2 2 O A dq dx V 4 2 2 d x dx ln x d x 2 2 d x 2 2 V ln x d x 4 2 2 2 2 V L L d d L ln 4 L ln 19

F( ) F( ) Induced Dipole Moment Many molecules such as H O have a permanent electric 2 2 2 dipole moment. These are known as "polar" molecules. For others, such as O, N, etc. the electric dipole moment is zero. These are known as "nonpolar" molecules. One such molecule is shown in fig. a. The electric dipole moment p is zero because the center of the positive charge coincides with the center of the negative charge. In fig. b we show what happens when an electric field E is applied to a nonpolar molecule. The electric forces on the positive and negative charges are equal in magnitude but opposite in direction. As a result the centers of the positive and negative charges move in opposite directions and do not coincide. Thus a nonzero electric dipole moment p appears. This is known as "induced" electric dipole moment, and the molecule is said to be "polarized." When the electric field is removed p disappears. 2

W qv Equipotential Surfaces A collection of points that have the same potential is known as an equipotential surface. Four such surfaces are shown in the figure. The work done by E as it moves a charge q between two points that have a potential difference V is given by W qv. For path I : W because V. I For path II: W because V. II For path III: W qv q V V. III 1 2 For path IV: W qv q V V. Note : IV 1 2 When a charge is moved on an equipotential surface V the work done by the electric field is zero: W. 21

A q F r E V B S The Electric Field E is Perpendicular to the Equipotential Surfaces Consider the equipotential surface at potential V. A charge q is moved by an electric field E from point A to point B along a path r. Points A and B and the path lie on S. Let's assume that the electric field E forms an angle with the path r. The work done by the electric field is: W Fr Frcos qercos. We also know that W. Thus: qercos, where q, E, r. Thus: cos 9. The correct picture is shown in the figure below. E S V 22

Examples of Equipotential Surfaces and the Corresponding Electric Field Lines Uniform electric field Isolated point charge Electric dipole Equipotential surfaces for a point charge q : q q V. Assume that V is constant r constant. 4 r 4 V Thus the equipotential surfaces are spheres with their center at the point charge and radius r q. 4 V 23

Calculating the Electric Field Efrom the Potential V Now we will tackle the reverse problem, i.e., determine E if we know V. Consider two equipotential surfaces that correspond to the values V and V dv separated by a distance ds as shown in the figure. Consider an arbitrary direction represented by the vector ds. We will allow the electric field to move a charge q from the equipotential surface V to the surface V dv. E s A B V V+dV The work done by the electric field is given by: W q dv (eq. 1). Also W Fdscos Eq dscos (eq. 2) If we compare these two equations we have: dv Eqds cos qdv Ecos. ds From triangle PAB we see that Ecos is the component E of E V along the direction s. Thus: Es. We have switched to the s V partial derivative symbols to emphasize that Es s involves only the variation of V along a spe cific s axis. V 24 s s

E s V V We have proved that Es. s s The component of E in any direction is the negative of the rate at which the electric potential changes with distance in this direction. V A B V+dV If we take s to be the x-, y-, and z-axes we get: V V V Ex ; Ey ; Ez x y z If we know the function V( x, y, z) we can determine the components of E and thus the vector E itself : V ˆ E i V ˆ j V kˆ x y z For the simple situation in which the electric field E is V uniform E, where s is perpendicular to the s equipotential surface. 25

y q 1 q 2 r 12 r 23 r 13 Potential Energy U of a System of Point Charges We define U as the work required to assemble the system of charges one by one, bringing each charge from infinity to its final position. Using the above definition we will prove that for a system of three point charges U is given by: O q 3 x U qq qq qq 4 r 4 r 4 r 1 2 2 3 1 3 12 23 13 Note : Each pair of charges is counted only once. i, j1 i j For a system of n point charges q the potential energy U is given by: U 1 4 n qq i r ij j. Here r is the separation between q and q ij The summation condition i j is imposed so that, as in the case of three point charges, each pair of charges is counted only once. i i j. 26

O y y O O y q 1 Step 3 q 1 Step 1 x Step 2 q 2 r 12 q 1 r 12 x q 2 x r 13 r 23 q 3 Step 1 : W 1 2 2 3 3 Bring in q : (no other charges around) Step 2 : W V(2) W V (3) qv(2) Step 3 : W 1 Bring in q : q qq W 4 4 qv(3) 2 1 1 2 2 12 r 12 r Bring in q : 1 4 q r 1 2 3 q r 13 23 1 qq 1 3 qq W3 4 r13 r23 W W W W 1 2 3 2 3 qq 1 2 qq 2 3 qq 1 3 4 r 4 r 4 r 12 23 13 27

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Energy of a system of charged conductors 29

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=1/4 (q 2 /L+q 3 /2L +q 4 /L) 31

conductor Potential of an Isolated Conductor We shall prove that all the points on a conductor A path E B (either on the surface or inside) have the same potential. A conductor is an equipotential surface. Consider two points A and B on or inside a conductor. The potential difference V B A between these two points is given by the equation: B V V EdS. B V A A We already know that the electrostatic field Thus the integral above vanishes, and V V for any two points on or inside the conductor. B E inside a conductor is zero. A 32

Isolated Conductor in an External Electric Field We already know that the surface of a conductor is an equipotential surface. We also know that the electric field lines are perpendicular to the equipotential surfaces. From these two statements it follows that the electric field vector E is perpendicular to the conductor surface, as shown in the figure. All the charges of the conductor reside on the surface and arrange themselves in such a way so that the net electric field inside the conductor E in. The electric field just outside the conductor is : Eout. 33

E ˆn E in E out E nˆ Electric Field and Potential in and around a Charged Conductor : A Summary ˆn 1. All the charges reside on the conductor surface. 2. The electric field inside the conductor is zero: E. 3. The electric field just outside the conductor is: Eout. 4. The electric field just outside the conductor is perpendicular to the conductor surface. 5. All the points on the surface and inside the conductor have the same potential. The conductor is an equipotential surface. in 34

Electric Field and Electric Potential for a Spherical Conductor of Radius and Charge q R 1 q For r R, V. 4 R 1 q For r R, V. 4 r For r R, E. 1 q For r R, E. 2 4 r R Note : Outside the spherical conductor, the electric field and the electric potential are identical to that of a point charge equal to the net conductor charge and placed 35 at the center of the sphr e e.