Chapter 25 Electric Potential Specific goals 1. Derive equations for the difference in potential energy, and the potential difference, between two points. 2. Define the potential difference between two points, the potential at a point, the volt, and the electron-volt. 3. Derive an expression for the potential difference between two points in a uniform electric field. 4. Find the potential at any point due to a group of point charges. 5. Define the electric potential energy due to a group of point charges. 6. Solve problems on the above. Electrical Potential Energy ⚫ When a test charge is placed in an electric field, it experiences a force ⚫ F = q E o ⚫ ⚫ ⚫ The force is conservative If the test charge is moved in the field by some external agent, the work done by the field is the negative of the work done by the external agent ds is an infinitesimal displacement vector that is oriented tangent to a path through space Section 25.1 Electric Potential Energy, cont ⚫ The work done by the electric field is F ds = qoE ds ⚫ As this work is done by the field, the potential energy of the charge-field system is changed by ΔU = −qoE ds ⚫ For a finite displacement of the charge from A to B, B U = UB − U A = −qo E ds A Section 25.1 Electric Potential Energy, final ⚫ ⚫ Because the force is conservative, the line integral does not depend on the path taken by the charge This is the change in potential energy of the system Section 25.1 Electric Potential ⚫ The potential energy per unit charge, U/qo, is the electric potential ⚫ The potential is characteristic of the field only ⚫ ⚫ ⚫ ⚫ The potential energy is characteristic of the charge-field system The potential is independent of the value of qo The potential has a value at every point in an electric field The electric potential is U V= qo Section 25.1 Electric Potential, cont. ⚫ The potential is a scalar quantity ⚫ ⚫ Since energy is a scalar As a charged particle moves in an electric field, it will experience a change in potential B U V = = − E ds A qo Section 25.1 Electric Potential, final ⚫ ⚫ ⚫ The difference in potential is the meaningful quantity We often take the value of the potential to be zero at some convenient point in the field Electric potential is a scalar characteristic of an electric field, independent of any charges that may be placed in the field Section 25.1 Work and Electric Potential ⚫ ⚫ Assume a charge moves in an electric field without any change in its kinetic energy The work performed on the charge is W = ΔU = q ΔV Section 25.1 Units ⚫ 1 V = 1 J/C ⚫ ⚫ ⚫ V is a volt It takes one joule of work to move a 1-coulomb charge through a potential difference of 1 volt In addition, 1 N/C = 1 V/m ⚫ This indicates we can interpret the electric field as a measure of the rate of change with position of the electric potential Section 25.1 Electron-Volts ⚫ ⚫ Another unit of energy that is commonly used in atomic and nuclear physics is the electron-volt One electron-volt is defined as the energy a charge-field system gains or loses when a charge of magnitude e (an electron or a proton) is moved through a potential difference of 1 volt ⚫ 1 eV = 1.60 x 10-19 J Section 25.1 Quick Quiz 25.1 Part I In the figure, two points A and B are located within a region in which there is an electric field. How would you describe the potential difference V = VB − VA? (a) It is positive. (b) It is negative. (c) It is zero. Section 25.1 Quick Quiz 25.1 Part II A negative charge is placed at A and then moved to B. How would you describe the change in potential energy of the charge–field system for this process? (a) It is positive. (b) It is negative. (c) It is zero. ΔU = q0·ΔV, so if a negative test charge is moved through a negative potential difference, the change in potential energy is positive. Work must be done to move the charge in the direction opposite to the electric force on it. Section 25.1 Potential Difference in a Uniform Field ⚫ The equations for electric potential can be simplified if the electric field is uniform: B B A A VB − VA = V = − E ds = −E ds = −Ed ⚫ The negative sign indicates that the electric potential at point B is lower than at point A ⚫ Electric field lines always point in the direction of decreasing electric potential Section 25.2 Energy and the Direction of Electric Field ⚫ When the electric field is directed downward, point B is at a lower potential than point A ⚫ When a positive test charge moves from A to B, the charge-field system loses potential energy Section 25.2 More About Directions ⚫ A system consisting of a positive charge and an electric field loses electric potential energy when the charge moves in the direction of the field ⚫ ⚫ An electric field does work on a positive charge when the charge moves in the direction of the electric field The charged particle gains kinetic energy equal to the potential energy lost by the charge-field system ⚫ Another example of Conservation of Energy Section 25.2 Directions, cont. ⚫ ⚫ If qo is negative, then ΔU is positive A system consisting of a negative charge and an electric field gains potential energy when the charge moves in the direction of the field ⚫ In order for a negative charge to move in the direction of the field, an external agent must do positive work on the charge Section 25.2 Equipotentials ⚫ ⚫ Point B is at a lower potential than point A Points B and C are at the same potential ⚫ ⚫ All points in a plane perpendicular to a uniform electric field are at the same electric potential The name equipotential surface is given to any surface consisting of a continuous distribution of points having the same electric potential Section 25.2 Charged Particle in a Uniform Field, Example ⚫ ⚫ ⚫ ⚫ ⚫ A positive charge is released from rest and moves in the direction of the electric field The change in potential is negative The change in potential energy is negative The force and acceleration are in the direction of the field Conservation of Energy can be used to find its speed Section 25.2 Quick Quiz 25.2 The labeled points in the figure are on a series of equipotential surfaces associated with an electric field. Rank (from greatest to least) the work done by the electric field on a positively charged particle that moves from A to B, from B to C, from C to D, and from D to E. B to C, C to D, A to B, D to E Moving from B to C decreases the electric potential by 2 V, so the electric field performs 2 J of work on each coulomb of positive charge that moves. Moving from C to D decreases the electric potential by 1 V, so 1 J of work is done by the field. It takes no work to move the charge from A to B because the electric potential does not change. Moving from D to E increases the electric potential by 1 V; therefore, the field does −1 J of work per unit of positive charge that moves. Section 25.2 Example 25.1: The Electric Field Between Two Parallel Plates of Opposite Charge A battery has a specified potential difference V between its terminals and establishes that potential difference between conductors attached to the terminals. A 12-V battery is connected between two parallel plates as shown in the figure. The separation between the plates is d = 0.30 cm, and we assume the electric field between the plates to be uniform. (This assumption is reasonable if the plate separation is small relative to the plate dimensions and we do not consider locations near the plate edges.) Find the magnitude of the electric field between the plates. Section 25.2 Example 25.1: The Electric Field Between Two Parallel Plates of Opposite Charge E= VB − VA d 12 V 3 = = 4.0 10 V/m −2 0.30 10 m Section 25.2 Example 25.2: Motion of a Proton in a Uniform Electric Field A proton is released from rest at point A in a uniform electric field that has a magnitude of 4 8.0 10 V/m, as shown in the figure. The proton undergoes a displacement of magnitude d = 0.50 m to point B in the direction of E. Find the speed of the proton after completing the displacement. Section 25.2 Example 25.2: Motion of a Proton in a Uniform Electric Field 1 2 mv − 0 + eV = 0 2 K + U E = 0 v= v= −2eV = m −2e ( − Ed ) m = 2eEd m 2 (1.6 10 −19 C )( 8.0 10 4 V ) ( 0.50 m ) 1.67 10 −27 kg = 2.8 106 m/s Section 25.2 Potential and Point Charges ⚫ A positive point charge produces a field directed radially outward B VB − VA = − E d s A E d s = ke q rˆ d s 2 r q E d s = ke 2 dr r ke q E= 2 r r rˆ d s = ds cos ds cos = dr dr q VB − VA = −ke q 2 = ke rA r r 1 1 VB − VA = ke q − rB rA rB VA = 0 at rA = q V = ke r Section 25.3 rB rA Potential and Point Charges, cont. ⚫ ⚫ ⚫ The electric potential is independent of the path between points A and B It is customary to choose a reference potential of V = 0 at rA = ∞ Then the potential at some point r is q V = ke r Section 25.3 Electric Potential with Multiple Charges ⚫ The electric potential due to several point charges is the sum of the potentials due to each individual charge ⚫ ⚫ This is another example of the superposition principle The sum is the algebraic sum ⚫ qi V = ke i ri V = 0 at r = ∞ Section 25.3 Potential Energy of Multiple charges ⚫ The potential energy of the system is . If the two charges are the same sign, U is positive and work must be done to bring the charges together. ⚫ If the two charges have opposite signs, U is negative and work is done to keep the charges apart. ⚫ Section 25.3 U with Multiple Charges, final ⚫ ⚫ If there are more than two charges, then find U for each pair of charges and add them For three charges: q1q2 q1q3 q2q3 U = ke + + r r r 13 23 12 ⚫ The result is independent of the order of the charges Section 25.3 Quick Quiz 25.3 Part I In the figure, take q2 to be a negative source charge and q1 to be a second charge whose sign can be changed. If q1 is initially positive and is changed to a charge of the same magnitude but negative, what happens to the potential at the position of q1 due to q2? (a) It increases. (b) It decreases. (c) It remains the same. Section 25.3 Quick Quiz 25.3 Part II In the figure, take q2 to be a negative source charge and q1 to be a second charge whose sign can be changed. When q1 is changed from positive to negative, what happens to the potential energy of the two-charge system? (a) It increases. (b) It decreases. (c) It remains the same. Section 25.3 Example 25.3: The Electric Potential Due to Two Point Charges As shown in the figure, a charge q1 = 2.00 C is located at the origin and a charge q2 = −6.00 C is located at (0, 3.00) m. (A) Find the total electric potential due to these charges at the point P, whose coordinates are (4.00, 0) m. q1 q2 VP = ke + r1 r2 VP = ( 8.988 109 N m 2 /C 2 ) 2.00 106 C −6.00 10 −6 C + 4.00 m 5.00 m = −6.29 103 V Section 25.3 Example 25.3: The Electric Potential Due to Two Point Charges (B) Find the change in potential energy of the system of two charges plus a third charge q3 = 3.00 C as the latter charge moves from infinity to point P. U f = q3VP U E = U f − U i = q3VP − 0 = ( 3.00 10−6 C )( −6.29 103 V ) = −1.89 10−2 J Section 25.3 Finding E From V ⚫ Assume, to start, that the field has only an x component dV Ex = − dx ⚫ Similar statements would apply to the y and z components ⚫ Equipotential surfaces must always be perpendicular to the electric field lines passing through them Section 25.4 E and V for an Infinite Sheet of Charge ⚫ ⚫ ⚫ The equipotential lines are the dashed blue lines The electric field lines are the brown lines The equipotential lines are everywhere perpendicular to the field lines Section 25.4 E and V for a Point Charge ⚫ ⚫ ⚫ The equipotential lines are the dashed blue lines The electric field lines are the brown lines The equipotential lines are everywhere perpendicular to the field lines Section 25.4 E and V for a Dipole ⚫ ⚫ ⚫ The equipotential lines are the dashed blue lines The electric field lines are the brown lines The equipotential lines are everywhere perpendicular to the field lines Section 25.4 Electric Field from Potential, General ⚫ ⚫ In general, the electric potential is a function of all three dimensions Given V (x, y, z) you can find Ex, Ey and Ez as partial derivatives V Ex = − x V Ey = − y V Ez = − z Section 25.4 Quick Quiz 25.4 Part I In a certain region of space, the electric potential is zero everywhere along the x axis. From this information, you can conclude that the x component of the electric field in this region is (a) zero, (b) in the positive x direction, or (c) in the negative x direction. Section 25.4 Quick Quiz 25.4 Part II Suppose the electric potential is +2 V everywhere along the x axis. From this information, you can conclude that the x component of the electric field in this region is (a) zero, (b) in the positive x direction, or (c) in the negative x direction. Section 25.4 Tutorial Problems NB: All problems are from the prescribed text book (9th Edition) P 25.1 P 25.5 P 25.14 P 25.25 P 25.27