Download Presentation
## Chapter 23

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**Chapter 23**Electric Potential Energy**Electric Potential (Voltage)**Electric potential is a scalar (not a vector). It can be positive or negative.**Potential of multiple charges**When you have more than one charge and you want to find the total potential, you simply add the potential due to each individual charge. q3 P q1 q2**y**P 3.00m x q2 q1 4.00m Point charge potential A 1.00 μC point charge is located at the origin and a second point charge of -4.00 μC is located on the x-axis. Find the total electric potential at P due to these charges.**y**P 3.00m x q2 q1 4.00m Point charge potential (The faster way) A 1.00 μC point charge is located at the origin and a second point charge of -4.00 μC is located on the x-axis. Find the total electric potential at P due to these charges.**y**P 3.00m S x q2 q1 4.00m Find Vs Point S is mid way between q1 and q2. Find the potential at S.**y**P 3.00m S x q2 q1 4.00m Potential DifferenceΔV FindΔVSP.**Why is potential useful?**q A B Say you want to move a charge q from A to B. In general there are two forces: Electric force (from the electric field) FE The force you apply in pushing the charge Fyou. It turns out knowing ΔV is sufficient in calculating the work involved in this process:**The meaning of work**Work means energy transfer. WE means energy transferred from the electric field to the charge. For example, if WE=100J, it means the electric field has transferred 100J of energy to the charge. If the charge is moving in free space (with no forces other than the electric force), then the energy gained by the charge becomes its KE. In real life applications, this energy from the electric field is extracted to power your appliances.**What about Wyou?**q A B**y**P 3.00m S x q2 q1 4.00m Final velocity Find thefinal velocityof a charge of 2μCand m = 1greleased at rest from P to S.**Energy released by a field**This is the energy released by the field, which could be used to power electrical appliances. In other words, energy can be extracted from a field when a charge moves from high potential to low potential.**Potential at Infinity**By convention, the electric potential at infinity is set to zero: Therefore, the change in KE for a charge released from infinity is given by:**y**P 3.00m S x q2 q1 4.00m Point charge potential What is the change in KE of a 3.00μC point charge moving from infinity to point P?**q2**r12 q1 PE for a pair of charges Like charges: U12> 0 Opposite charges: U12 < 0 Do not confuse potential energy with potential! Units: Potential Energy: J (1J=1VC) Potential: V**r12**q1 q2 r13 r23 q3 Multiple Charges Total potential energy obtained by calculating Ufor every pair of charges and add them up. In this case there are total of three pairs:**How many pairs do you see?**q1 q2 q3 q4**y**P 3.00m S x q2 q1 4.00m POTENTIAL ENERGY 1 q3**y**P 3.00m S x q2 q1 4.00m POTENTIAL ENERGY 2 q3**The flow of charge**If unrestrained: Positive charge flow from highV to lowV Negative charge flow from lowV to highV High V E field Low V + −**Obtaining E from V**The component of the electric field in a particular direction is the negative of the gradient of the potential in that direction. E field always point from high potential to low potential.**Example**If V = A(3xy2 - 2yz2) where A=5 volts/m3, find the components ofE.**Example: electric Dipole**P + -**V FROM E: PARALLEL PLATES (UNIFORM E FIELD)**V decreases along the direction of the E field. E field points in the direction of decreasing V.**V from E**A B**Special case**A B**E from V**A B**Spherically symmetric case**For spherically symmetric charge distributions, the electric fields are radial and we have: The above equations are often combined with the convention of V∞=0 to give the potential:**Potential of a point charge**Integrate from infinity to P P q**Tricks to get V from E**A B Sometimes the electric field changes from one region to the next. In this case, the integral for ΔV breaks into several terms.**Potential of a uniform spherical charge distribution**Suppose the sphere has a uniform volume charge density ρ, find potential inside and outside the sphere. R**Case 1: Outside the sphere (r > R)**Integrate from infinity to P P R**Case 2: Inside the sphere (r < R)**Integrate from infinity to P P R**Equipotential surfaces**• They are defined as a surface in space on which the potential is the same for every point (surfaces of constant voltage) • The electric field at every point of an equipotential surface is perpendicular to the surface**Equipotential lines perpendicular to E field**Consider the equation: To get zero, all line elements dsalong the trace must be perpendicular to E.**Electric Potential of a Charged Conductor**• E =0 inside conductor in equilibrium. • Any net charge resides on the surface of the conductor. • Electric field immediately outside the conductor is perpendicular to the surface of the conductor.**dq**r a x x P Continuous charge distribution Find the electric potential at a point P located on the axis of a uniformly charged ring of radius a and a total charge Q. The plane of the ring is perpendicular to the x-axis. Each charge element on the ring is the same distance from P, so r is a constant in the integration: