The Electric Potential Lecture 7 The Electric Potential

Electric Potential Energy The electric force, like the gravitational force, is a conservative force. (Conservative force: The work is path-independent.) As in mechanics, work is Work done on the positive charge by moving it from A to B A B d

The work done by a conservative force equals the negative of the change in potential energy, DPE This equation is valid only for the case of a uniform electric field If a charged particle moves perpendicular to electric field lines, no work is done. if d  E

Question: In the figure, a proton moves from point i to point f in a uniform electric field directed as shown. Does the electric field do positive, negative or no work on the proton? A: positive B: negative C: no work is done on the proton Solution:

Electric potential is a scalar quantity The potential difference between points A and B, VB-VA, is defined as the change in potential energy (final minus initial value) of a charge, q, moved from A to B, divided by the charge Electric potential is a scalar quantity Electric potential difference is a measure of electric energy per unit charge Potential is often referred to as “voltage” If the work done by the electric field is zero, then the electric potential must be constant

Electric potential difference is the work done to move a charge from a point A to a point B divided by the magnitude of the charge. Thus the SI units of electric potential difference In other words, 1 J of work is required to move a 1 C of charge between two points that are at potential difference of 1 V Question: How can a bird stand on a high voltage line without getting zapped?

Units of electric field (N/C) can be expressed in terms of the units of potential (as volts per meter) Because the positive tends to move in the direction of the electric field, work must be done on the charge to move it in the direction, opposite the field. Thus, A positive charge gains electric potential energy when it is moved in a direction opposite the electric field A negative charge looses electrical potential energy when it moves in the direction opposite the electric field

Example : A uniform electric field of magnitude 250 V/m is directed in the positive x direction. A +12µC charge moves from the origin to the point (x,y) = (20cm, 50cm). (a) What was the change in the potential energy of this charge? (b) Through what potential difference did the charge move? Begin by drawing a picture of the situation, including the direction of the electric field, and the start and end point of the motion. (a) The change potential energy is given by the charge times the field times the distance moved parallel to the field. Although the charge moves 50cm in the y direction, the y direction is perpendicular to the field. Only the 20cm moved parallel to the field in the x direction matters for determining the change of potential energy. PE = -qEd = -(+12µC)(250 V/m)(0.20 m) = -6.0×10-4 . (b) The potential difference iS the difference of electric potential,. V = PE / q = -6.0×10-4 J / 12µC = -50 V.

Analogy between electric and gravitational fields The same kinetic-potential energy theorem works here If a positive charge is released from A, it accelerates in the direction of electric field, i.e. gains kinetic energy If a negative charge is released from A, it accelerates in the direction opposite the electric field A A d d q m B B

Example: motion of an electron What is the speed of an electron accelerated from rest across a potential difference of 100V? Given: DV=100 V me = 9.11´10-31 kg mp = 1.67´10-27 kg |e| = 1.60´10-19 C Find: ve=? vp=? Vab

+ - Problem: A proton is placed between two parallel conducting plates in a vacuum as shown. The potential difference between the two plates is 450 V. The proton is released from rest close to the positive plate. What is the kinetic energy of the proton when it reaches the negative plate? Example (a) Through what potential difference would an electron need to accelerate to achieve a speed of 60% of the speed of light, starting from rest? (The speed of light is 3.00×108 m/s.) (a) The final speed of the electron is vf = 0.6(3.00×108 m/s) = 1.80×108 m/s. At this speed, the energy (non-relativistic) is the kinetic energy, E = ½mvf² = 0.5(9.11×10-31 kg)(1.80×108 m/s)² = 1.48×10-14 J. This energy must equal the change in potential energy from moving through a potential difference, V, E = PE = qV. Therefore: V = E/q = (1.48×10-14 J)/(-1.6×10-19 C) = -9.25×104 V.

Electric potential and potential energy due to point charges Electric circuits: point of zero potential is defined by grounding some point in the circuit Electric potential due to a point charge at a point in space: point of zero potential is taken at an infinite distance from the charge With this choice, a potential can be found as Note: the potential depends only on charge of an object, q, and a distance from this object to a point in space, r.

Superposition principle for potentials If more than one point charge is present, their electric potential can be found by applying superposition principle The total electric potential at some point P due to several point charges is the algebraic sum of the electric potentials due to the individual charges. Remember that potentials are scalar quantities!

Potential energy of a system of point charges Consider a system of two particles If V1 is the electric potential due to charge q1 at a point P, then work required to bring the charge q2 from infinity to P without acceleration is q2V1. If a distance between P and q1 is r, then by definition Potential energy is positive if charges are of the same sign and vice versa. q2 q1 r P A

Example: potential energy of an ion Three ions, Na+, Na+, and Cl-, located such, that they form corners of an equilateral triangle of side 2 nm in water. What is the electric potential energy of one of the Na+ ions? Cl- ? Na+ Na+

Potentials and charged conductors Recall that work is opposite of the change in potential energy, No work is required to move a charge between two points that are at the same potential. That is, W=0 if VB=VA Recall: all charge of the charged conductor is located on its surface electric field, E, is always perpendicular to its surface, i.e. no work is done if charges are moved along the surface Thus: potential is constant everywhere on the surface of a charged conductor in equilibrium … but that’s not all!

Because the electric field in zero inside the conductor, no work is required to move charges between any two points, i.e. If work is zero, any two points inside the conductor have the same potential, i.e. potential is constant everywhere inside a conductor Finally, since one of the points can be arbitrarily close to the surface of the conductor, the electric potential is constant everywhere inside a conductor and equal to its value at the surface! Note that the potential inside a conductor is not necessarily zero, even though the interior electric field is always zero!

The electron volt A unit of energy commonly used in atomic, nuclear and particle physics is electron volt (eV) The electron volt is defined as the energy that electron (or proton) gains when accelerating through a potential difference of 1 V Relation to SI: 1 eV = 1.60´10-19 C·V = 1.60´10-19 J Vab=1 V

Example : ionization energy of the electron in a hydrogen atom In the Bohr model of a hydrogen atom, the electron, if it is in the ground state, orbits the proton at a distance of r = 5.29´10-11 m. Find the ionization energy of the atom, i.e. the energy required to remove the electron from the atom. Note that the Bohr model, the idea of electrons as tiny balls orbiting the nucleus, is not a very good model of the atom. A better picture is one in which the electron is spread out around the nucleus in a cloud of varying density; however, the Bohr model does give the right answer for the ionization energy

In the Bohr model of a hydrogen atom, the electron, if it is in the ground state, orbits the proton at a distance of r = 5.29 x 10-11 m. Find the ionization energy, i.e. the energy required to remove the electron from the atom. Given: r = 5.292 x 10-11 m me = 9.11´10-31 kg mp = 1.67´10-27 kg |e| = 1.60´10-19 C Find: E=? The ionization energy equals to the total energy of the electron-proton system, with The velocity of e can be found by analyzing the force on the electron. This force is the Coulomb force; because the electron travels in a circular orbit, the acceleration will be the centripetal acceleration: or or Thus, total energy is

Calculating the Potential from the Field To calculate the electric potential from the electric field we start with the definition of the work dW done on a particle with charge q by a force F over a displacement ds In this case the force is provided by the electric field F = qE Integrating the work done by the electric force on the particle as it moves in the electric field from some initial point i to some final point f we obtain

Calculating the Potential from the Field Remembering the relation between the change in electric potential and the work done … …we find Taking the convention that the electric potential is zero at infinity we can express the electric potential in terms of the electric field as ( i = , f = x)

Example - Charge moves in E field Given the uniform electric field E, find the potential difference Vf-Vi by moving a test charge q0 along the path icf. Idea: Integrate Eds along the path connecting ic then cf. (Imagine that we move a test charge q0 from i to c and then from c to f.)

Example - Charge moves in E field distance = sqrt(2) d by Pythagoras

Question A: 0 B: -Ed C: +Ed D: -1/2 Ed We just derived Vf-Vi for the path i -> c -> f. What is Vf-Vi when going directly from i to f ? A: 0 B: -Ed C: +Ed D: -1/2 Ed Quick: DV is independent of path. Explicit: DV = -  E . ds =  E ds = - Ed

Examples & Problems W = Δk = kf - ki W = 0-ki q V = -1/2 mv2 Example 1: What potential difference is needed to stop an electron with an initial speed of 4.2*105m/s? W = Δk = kf - ki W = 0-ki q V = -1/2 mv2 V = -(1/2 mv2) / q V = -(1/2 x 9.1x10-31x(4.2x105)2)/1.5x10-19 V = -0.57 volt Example 2: An ion accelerated through a potential difference of 115V experiences an increase in potential energy of 7.37*10-17J. Calculate the charge on the ion. VB - VA = Wab/ q q = Wab/ VB - VA q= 6.41x10-19C

r = 2 m E = 1.0´10-7/(2x8.85x10-12)= 5650 N/C V = Ed d = V/E Example 3: An infinite charged sheet has a surface charge density s of 1.0*10-7 C/m2.  How far apart are the equipotential surfaces whose potentials differ by 5.0 V?                    E = 1.0´10-7/(2x8.85x10-12)= 5650 N/C V = Ed d = V/E d = 8.8x10-4m Example 4: At what distance from a point charge of 8mC would the potential equal 3.6*104V?   r = 2 m

Example 5: At a distance r away from a point charge q, the electrical potential is V=400v and the magnitude of the electric field is E=150N/C.  Determine the value of q and r. r = 2.67m

E = F/q V = (F/q) d V = (1.6x1015/1.6x10-19) x 0.05 = 5x1032v Example 6: Calculate the value of the electric potential at point P due to the charge configuration shown in Figure 5.19. Use the values q1=5mC, q2=-10mC, a=0.4m, and b=0.5m.                                                                   V = V1 + V2 + V3 + V4 Example 7: Two large parallel conducting plates are 10cm apart and carry equal and opposite charges on their facing surfaces.  An electron placed midway between the two plates experiences a force of 1.6*1015N.  What is the potential difference between the plates? V=Ed E = F/q V = (F/q) d V = (1.6x1015/1.6x10-19) x 0.05 = 5x1032v

Example 8: Two point charges are located as shown in, where ql=+4mC, q2=-2mC, a=0.30m, and b=0.90m. Calculate the value of the electrical potential at points P1, and P2.  Which point is at the higher potential?                                                                     Vp1 = V1 + V2

Example 9:  In figure 5.22 prove that the work required to put four charges together on the corner of a square of radius a is given by  (w=-0.21q2/eoa). U = U12 + U13 + U14 + U23 + U24 + U34                                                                                                                                                           

Example Assume we have a system of three point charges: q1 = +1.50 C q2 = +2.50 C q3 = -3.50 C. q1 is located at (0,a) q2 is located at (0,0) q3 is located at (b,0) a = 8.00 m and b = 6.00 m. What is the electric potential at point P located at (b,a)?

r1 The electric potential at point P is given by the sum of the electric potential from the three charges r2 r3

(1) Two charges q=+2*10-6C are fixed in space a distance d=2cm apart, as shown in figure (a) What is the electric potential at point C? (b) You bring a third charge qn=2.0*10-6C very slowly from infinity to C.  How much work must you do?  (c) What is the potential energy U of the configuration when the third charge is in place? (2) Four equal point charges of charge q=+5mC are located at the corners of a 30cm by 40cm rectangle. Calculate the electric potential energy stored in this charge configuration. (3) Two point charges, Q1=+5nC and Q2=-3nC, are separated by 35cm. (a) What is the potential energy of the pair?  What is the significance of the algebraic sign of your answer? (b) What is the electric potential at a point midway between the charges?

(4) What is the potential at the center of the square shown in figure 5.9? Assume that   q1= +1 ´10-8C,  q2= -2´10-8C,  q3=+3´10-8C, q4=+2´10-8C, and a=1m. (5) Two charges of 2mC and -6mC are located at positions (0,0) m and (0,3) m, respectively as shown in figure.  (i) Find the total electric potential due to these charges at point (4,0) m.  (ii) How much work is required to bring a 3mC charge from ¥ to the point P?  (iii) What is the potential energy for the three charges?

(6) A particle having a charge q=3 (6) A particle having a charge q=3*10-9C moves from point a to point b along a straight line, a total distance d=0.5m.  The electric field is uniform along this line, in the direction from a to b, with magnitude E=200N/C.  Determine the force on q, the work done on it by the electric field, and the potential difference Va-Vb. (7) For the charge configuration shown in figure , Show that V(r) for the points on the vertical axis, assuming r >> a, is given by