Electric Potential. Slide 1 / 29. Slide 2 / 29. Slide 3 / 29. Slide 4 / 29. Slide 6 / 29. Slide 5 / 29. Work done in a Uniform Electric Field

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1 Slie 1 / 29 Slie 2 / 29 lectric Potential Slie 3 / 29 Work one in a Uniform lectric Fiel Slie 4 / 29 Work one in a Uniform lectric Fiel point a point b The path which the particle follows through the uniform electric fiel is inepenent of the path it takes. To unerstan this, think back to gravitational potential energy. Two Hikers of equal mass walk up a hill. Hiker takes the shortest path to get to the top of the hill an Hiker takes the longest route to the top. In the en which has a greater potential energy? The answer is they have the same potential energy. The force is constant in a uniform magnetic fiel so we can use: We iscusse in previous chapters that: an When we sub in into the force equation we get: H Hiker Hiker oth hikers will have a potential of mgh Therefore the work one in moving a particle through a uniform electric fiel is: or Slie 5 / 29 Slie 6 / 29 1 Two charges of charge Q an -2Q are separate by a istance. The work require to move a test charge of magnitue Q from point to point which is miway between the two charges is epenent on the path inversely proportional to the istance epenent on the isplacement zero more information is require Q -Q 2 negative charge of coulombs is palce in an upwar irecte uniform electric fiel of. When the charge is move 2 meters upwar, the work one by the electric force on the charge is 1.8 x 10 4 J x 10 4 J 1.8 x J x J 3.6 x 10 4 J

2 Slie 7 / 29 Uniform lectric Fiel We sai before that an uniform electric fiel is similar to a case uner the force of gravity. When we have an object of mass m, an move it up an own we change the amount of potential energy it has. When we go against the force of gravity the potential energy increases an when we allow gravity to take over it falls to a lower potential. Slie 8 / 29 lectric Potential nergy of Two Point harges For electric potential between two charges we will have to eal with two ifferent cases. The first eals with a test charge moving away from another charge linearly an the secon case is when the test charge moves raially away from the other charge. This hols true in a uniform electric fiel. When we oppose the force acting on the particle within the fiel we will increase the particles potential energy an if we allow it to move in the irection of the force it falls to a lower potential. ase 1: r p 2 Since the force ue to this system is not constant we have to use the integral representation of work F F Q p 1 r a r b Slie 9 / 29 Slie 10 / 29 lectric Potential nergy of Two Point harges Now if we eal with the more general case of moving the test charge raially away from the other charge we will fin that the potential energy is not epenent on the path we take. r l From the sketch we see that: Q Slie 11 / 29 lectric Potential nergy with several point charges In establishing the formations of the charges in the previous example there woul also be a potential energy. Slie 12 / 29 3 Three equal test charges of magnitue are place at the corners of a square. test charge of magnitue is place at the last open corner. What is the electric potential energy of the test charge? q J Test harge Q 1 q1 The equation for this is also the algebraic sum of the interactions between each of the charges. q J 27.8 J 16.4 J 1 J Q 3.5m Q 2 In this equation i<j because a charge oes not interact with itself, this way the equation takes into account all the interactions between the charges.

3 Slie 13 / 29 Slie 14 / 29 lectric Potential V ecreases V ecreases V increases - V increases From these examples we can say that when we move in the irection of the electric fiel the electric potential ecreases an as we move in the opposite irection the electric potential increases. Slie 15 / 29 lectric Potential Once again if there is a collection of charges then the net electric potential at some point p is equal to the algebraic sum of each of the charges electric potentials at that point. Once again if there is a collection of charges then the net electric potential at some point p is equal to the algebraic sum of each of the charges electric potentials at that point. q2 Slie 16 / 29 lectric Potential If a charge is istribute over a surface such as those in a coulomb's problem we break the overall object into small pieces of charge q. The net electric potential at a certain point is given by: (lectric Potential ue to a charge istribution) efore we iscusse that the work one by moving a charge through an electric fiel is given as: q1 an P q3 i is the subscript for the charge an the istance of the charge from the point were you are evaluating the net electric potential. Slie 17 / 29 Slie 18 / 29 4 Two charges Q1 an Q2 are place a istance apart from each other along a line an have charges 2Q an -Q respectively. t how many points is the electric potential zero. 5 What is the electric potential at point p? Q 1 = 3x10-12 Q 2 = 4x volts Q 1 none one two three 2Q -Q.9 volts 1.44 volts 1.8 volts 2.6 volts 3 cm Q 2 4 cm P more information is require

4 Slie 19 / 29 alculating lectric Potential for harge istributions (Sphere) If a soli metal sphere of raius is given a charge Q. What are the graphs of the electric fiel an the electric potential for the sphere. Slie 20 / 29 alculating lectric Potential for harge istributions (Thin onucting yliner ) First we have to recognize that since it is a conuctor all of the charge will resie on the surface of the sphere. When we apply Gauss's Law we fin the electric fiel insie the sphere to be zero, an the electric fiel outsie of the sphere to be the same for a point charge. (lectric Fiel of harge yliner) since =0, V a-v b=0. Therefore V a=v b an every point within the sphere is at the same electric potential. V Slie 21 / 29 alculating lectric Potential for harge istributions (Thin onucting yliner ) Slie 22 / 29 6 What is the potential energy of a positively charge ring of raius a a istance away from its center? To calculate the electric potential for other shapes you simply use a ifferent form of the equation like in oulomb's Law, where you break it into tiny pieces an a up the charges. a Slie 23 / 29 quipotential Surfaces n equipotential surface is a 3 imensional structure on which the electric potential is constant. For a sphere the equipotential surfaces woul be spheres, an to sketch the magnitue of each of the surfaces we use contour lines like on a topographic map, a map that shows changes in elevation. When the lines are getting further apart they inicate that the potential is ecreasing, when the are evenly space the potential is constant, an when they are getting closer together the potential is increasing. This same reasoning can be applie to more complicate shapes, in turn each of there equipotential surfaces woul be the same as the actual surface. V/4 V/9 V/16 Slie 24 / 29 7 test charge of magnitue q moves from point to point. What is the work one on the particle? 0 20q 10q 40q 50q - - V

5 Slie 25 / 29 8 test charge of magnitue q moves from point to point. What is the work one on the charge? Slie 26 / 29 9 test charge of magnitue q moves from point to point. What is the work one on the test charge? -10q 0 10q 30q 60q q -20q 40q 30q 10q - - Slie 27 / 29 Slie 28 / 29 Potential Graient The electric potential is the negative graient of the electric fiel meaning at each point it inicates the irection in which v ecreases most rapily with a change in its position. Therefore as we explaine before is in the irection in which the potential ecreases most rapily an is perpenicular to the equipotential surface. Slie 29 / charge ring of raius a has a potential energy given by the equation. What is the equation of its electric fiel?