PHY 184 Week 4 Notes-Electric Potential
PHY 184 Week 4 Notes-Electric Potential PHY 184
Popular in Physics for Scientists and Engineers II
Popular in Physics
This 5 page Class Notes was uploaded by Cameron Blochwitz on Saturday September 24, 2016. The Class Notes belongs to PHY 184 at Michigan State University taught by Oscar Naviliat Cuncic in Fall 2016. Since its upload, it has received 8 views. For similar materials see Physics for Scientists and Engineers II in Physics at Michigan State University.
Reviews for PHY 184 Week 4 Notes-Electric Potential
Report this Material
What is Karma?
Karma is the currency of StudySoup.
Date Created: 09/24/16
PHY 184 Week 4 Notes-Electric Potential 9/19-9/22 We have seen that gravity and electromagnetism are very similar in their behavior even going so far as to have almost identical equations 2 o Both have forces that are determined by 1/r These forces are all conservative forces Doesn’t depend on path taken o Work is independent Consider an electrostatic force acting in charge o The work is produced between i and f The workfintegral is ´ i F∙ds The variation of potential energy, ΔU, is the negative of work done by the electrostatic force o It is also the work done against the electrostatic force Like gravity we need to select a reference point is defined Potential is 0 when the charges are infinitely far apart The variation of the Potential Energy reads like ∆ U=U −f=U−W e,inf o o The negative shows Positive work for U<0 Negative work for U>0 Consider a charge q located in an electric field moving on d where d is at an angle θ to the electric field Force on the charge ´ ´ o F=q∙E W=F ∙d´ o Work done is ´ o W=qE∙d=qEd cosθ If displacement is along the direction of the field o Positive charge loses potential energy If displacement is opposite the field o Positive charge gains potential energy Electric Potential Electric potential is defined to be as the “electric potential energy per unit charge” V= U o q It describes a property of space and therefore is a scalar The unit of electric potential is the Volt, V 1 J o 1V= 1C Therefore, the units of an electric field are able to be given in V/m Variation of Electric Potential V f V i ∆U −W e ∆ V=V fV =i q − q= q = q ∆U f ∆ V= =− ∫∙d ´ s q i Energy of a Proton A Proton is between 2 plates o Potential difference is 150V o Proton is released from positive plate What is the K pt the negative plate? o We are able to use energy conservation to solve this problem Solving for gives Kf=−q∆V Acceleration though a system is able to be calculated again from energy conservation o The particle starts at rest so we can calculate acceleration in two stages ∆ K=|∆U = q∆V |+| 2=K K=e∆V +6 e∆V =7e∆V We are then able to insert numerical values o We are able to use the electron-Volt, eV, instead of a joule -12 1 eV is 1.602 x 10 Joules When an electron “falls” across a 1.5 Volt difference the energy gained by that electron is 1.5 eV Equipotential Lines-Gravity Consider a landscape with mountains o On a topographical map they appear as a series of rings set at a certain height All these lines have the same gravitational potential They are equipotential lines! Equipotential Surfaces An electric charge generates a field in space o Electric potential has a value everywhere in space o Those points that have the same electric potential form a surface o Charges that move along these surfaces have no change in potential energy When moving along Equipotential Surfaces there is no work done by the electric force Conversely, if the line integral of the field is 0 the electric potential must be constant o Therefore, moving a charge perpendicular to the force there is no work The charge moves along equipotential lines Any conductor forms an equipotential surface because the electric force is always perpendicular to the surface Finding Potential Difference in a Uniform Field f ∆ V=− E∙d´ s ∫i Preforming this integral gives the solution ∆ V=−Ed o If at position z 00 and the potential is V we hove o ∆ V=V (z−V =0E(z−z ) 0 o V z)=V 0Ez If potential at infinity is 0 f V r)−V ∞=V (r=− E∙d´ s o ∫i Projection away from charge along field V r)= kq o r Distance of Closest Approach A charge is fixed at x=0. A particle of mass m and charge q is 2 fired with an initial velocity v tow0rd the charge at from a distance d . 1 kq1 o V ( ) x kq q o U ( )q 2 x( ) 1 2 x Using energy conservation, we are able to find that mv2 1 = 1+ o o d f di 2 k 1 2 System of Charges The potential of a system of charges is given by superposition Minimum Potential Given 2 positive charges q and q a1 what p2sition is electric potential minimized Find V, set derivative to 0 and solve o Solving the equation gives x= r2 q2+1 √ q1 Potential Energy of a System In a system of charges each charge has equal weight If two charges have the same sign potential is positive, if opposite negative For a system greater o Consider every pair in the system and sum Electric Field from Electric Potential We have seen so far how to calculate the electric potential given the electric field It is important to know that we can go the other way too, using potential to get the electric field This is done using partial derivatives E= −∂V ,−∂V ,−∂V =−∇V o ( ∂x ∂y ∂z ) o If we have the potential, we are able to derive the electric field at any point Continuous Charge Distribution For a system of charges, the potential is just the sum of the potentials of the charges in the system To determine the electric potential due to a continuous distribution we divide the charge into charge elements and treat them as point charges dq o V pk ∫ r To find the potential at a point in a semicircle o We know that there is a symmetry giving us the variable θ o The length can be written as dL=Rdθ o Electric charge can be written dq=λdL Solving the integral gives V=πλk If we do the same for a Linear horizontal charge with length 2a and distance d 2e 2ind V=kλln √a +d +a o √a +d −a