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As a mass is raised above the surface of the earth, its potential energy increases. If we do not move too high above the surface of the earth, the gravitational field is constant, and the force of gravity remains constant. In a similar way, when a point charge is moved from one charged plate to another so that an external force is required, the electric field is constant and the force acting on the charge is constant. However, the potential energy increases as the charged object moves away from the plate that is attracting it. The change in potential energy of the charge as it moves from one height, h4, above a plate to another height, h,, can be determined by APE = FAh. This is equivalent to determining the area under a force-position graph. The change in electric potential energy of a charge g in moving from A to B is equal to the work required to move g from A to B against the electric field. W = APE = PEg - PE, If the particle is released, its potential energy will be changed to kinetic energy: gF 1 mv* - The standard unit for both energies is the joule (J). The speed of the particle can be calculated if the kinetic energy and the mass of the particle are known. The electric potential V at a given point is the electrical potential energy, PE, of a test charge g, situated at that point divided by the charge itself: PE 9. The Sl unit of electric potential is the joule/coulomb also called the volt (V). The potential difference between these two points is simply the difference between potential V, and V4. The change in electric potential energy of a charge g in moving from A to B can be expressed as: APE = qVg - qVa=q(Vg - Va) = qAV. We can define the electron volt as the energy acquired by an electron as a result of moving through a potential difference of 1V and 1 eV is equal to (1.602 x 10719 C)(1V) = 1.602 x 10719 J. The relationship between the electric field and the potential difference AV between the points "a" and "b" separated by the distance dis AV = Ed. An equipotential line is one where all points on that line are at the same potential. That is, the potential difference between any two points on the line is zero. Because of this, no work is required to move a charge from one point to another. An equipotential line must be perpendicular to the electric field at any point. For two parallel plates, the electric field lines are perpendicular to the plates, while the lines of equipotential are parallel to the plates. Question(s): 1. The physics of moving a small charge through a potential difference. How much work is needed to move a -8.0 uC charge from ground ("ground" in electrical terms means a potential of O V) to a point whose potential is +75 V? 2. The physics of an electron falling through a potential difference. How much kinetic energy will an electron gain (joules) if it falls through a potential difference of 350 V? 3. The physics of an electron moving from one plate to another. An electron acquires 4.2 x 10716 J of kinetic energy when it is accelerated by an electric field from Plate A to plate B. What is the potential difference between the plates, and which plate is at a higher potential? EaCalculating Electric Potential and Potential Difference From the example considered previously, we can now calculate the electric potential at different places, and the potential difference. F (force needed to pull charge up) electric field PE, || g Fr T 4E (electrical force pulling down) h |, negative plate We determined that the electric potential energy at the first height was 0.100 J. The electric potential at this point is y-PE 010000000y g, 10.0C The electric potential energy at the second height was 0.400 J. The electric potential at this point is P 0.400J V. :_E2 = w ) 0.0400 V g, 10.0C The potential difference between these two points is AV=V,-V,=0.0400V-0.0100 V =0.0300 V The change in electric potential energy of a charge q in moving from A to B is equal to the work required to move g from A to B against the electric field. This can now be expressed as follows. APE =gV - qVa =q(V3 - Va) = gAV