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Test 1 review

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1 review
by: Katie Hathaway

Test 1 review PHYS 202

Katie Hathaway
GPA 3.9
General Physics II

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About this Document

Review of major concepts and homework problem set answers
General Physics II
Study Guide
50 ?




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1 review
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April D

Popular in General Physics II

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This 10 page Study Guide was uploaded by Katie Hathaway on Friday September 18, 2015. The Study Guide belongs to PHYS 202 at University of South Carolina taught by in Fall 2015. Since its upload, it has received 122 views. For similar materials see General Physics II in Physics 2 at University of South Carolina.


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Date Created: 09/18/15
Law of Charges Like charges repel and opposite charges attract Thus electric charge is a fundamental property of the elementary particles electrons protons neutrons out of which atoms are made of Remember protons are positive and electrons are negative Objects do not seem to carry a net charge because they have equal amounts of positive and negative charge electrically neutral The unit of charge is the coulomb abbreviated C We can de ne the unit of charge in terms of the charge of one electron e 1602 x 103919 C In conductors one or more of the outermost electrons of the constituent atoms become free and move thoughout the solid There are two ways to charge an object o Conduction lnducUon What is the force between an electrically charged object q and a neutral object 0 It is always attractive due to polarization electronsprotons of object move Coulomb s law Consider two charges q1 and q2 placed at a distance r The two charges exert a force on each other that has the following characteristics 0 The force acts along the line connecting the two charges 0 The force is attractive for charges of opposite sign The force is repulsive for charges of the same sign 0 The magnitude of the force known as Coulomb force is given by the equation F 14nao q1q2r2 The constant so is known as the permittivity constant so 885 x 103912 Nm2C2 The Coulomb force has the same form as Newton39s gravitational force The net electric force exerted by a group of charges is equal to the vector sum of the contribution from each charge For example the net force exerted on by and is equal to F1 F12 F13 Remember these values are vectors magnitude and direction and so vector addition must be used The gravitational force that a uniform shell of mass m2 exerts a particle of mass m1 that is outside the shell is given the equation Because of the similarity between Newton39s gravitational law and Coulom39s law the same is true for the electric force exerted by a spherical shell of charge 02 on a point charge 01 A eld is a physical entity that extends throughout a volume of space and exerts forces The electric eld extends throughout space and exerts forces on charged particles If we place a positive point charge in an electric eld there will be a vector force on that charge in the direction of the electric eld with magnitude depending on strength of the eld Vector E vector Fqo units NC It is the surrounding charges that create the electric eld at a given point The Electric eld DOES NOT depend upon the test charge At point P a distance r from q we place the test charge qo is equal to F 14TEEo CICIol39z I EFCI0 14TEEo CICIoCIOFZ 14TEEo Clr2 Suppose we have many charges the electric eld at any point in space will have contributions from all the charges The electric eld at any point in space is the superposition of the electric eld from n charges is E E1 E2 E3 En Electric Field Lines At any point P the electric eld vector E is tangent to the electric eld lines and the magnitude of the electric eld is proportional to the density of the electric eld lines We can represent the electric eld graphically by electric eld lines curves representing vector force exerted on test charge Electric eld lines will originate on positive charges and extend away from then terminate on negative charges The electric force at a given point in space is tangent to the electric eld line through that point We can use the superposition principle to calculate the electric eld from two point charges Enet E E If the eld lines connect we have an attractive force If the eld lines seem to spread out we have a repulsive force Strongest in middle unless equal then equals zero Guass Law Electric ux density of electric eld lines through an area A the electric ux I EAcosG The angle 6 is between the electric field and the area vector Vector A points normal to the surface and has a length A easiest way to understand is as the integral of E and dA The ux of vector E through any closed surface is multiplied my so is equal to the net charge of qenc enclosed by the surface Eo q Gauss law holds true for any closed surface Net charge accounts for sign of each charge lgnore charges outside the surface Gauss39 law and Coulomb39s law are different ways of describing the relation between electric charge and electric eld in static cases One can derive Coulomb39s law from Gauss law and vice versa The electric eld inside of a conductor equals zero all charges reside on the conductor surface There is no charge on the cavity walls all excess charge q on outer surface The conductor shields any charge within it from electric elds created outside the conductor The electric eld just outside the surface of a conductor is perpendicular to the surface at equilibrium under electrostatic conditions EOEo I EA q so l EqA 1 so cosO1 so omitted The ratio oqA is known as surface charge density l E 0 so A sharp point will have eld lines much closer together so the eld is much stronger and looks much more like the eld of a point charge Use a quotcharge densityquot to describe the distribution of charge which will different depending on the geometry A charge per length Cm 0 charge per area Cm2 p charge per volume Cm3 Charged rod using symmetry electric eld vector points radially outwards and has same magnitude for points at distance r from rod Use Gaussian surface S a cylinder with radius r and height h divide S into 3 sections 1 and 2 equal 0 because electric eld at right angles with normal surface I 2nrhEcosO 2nrhE From gauss law I q so Ah so 2nrhE Ah sol E A2nr so Uniform charged sheet the electric eld vector E is perpendicular to sheet and points away Cylindrical surface S with the caps of area A on either side of sheet divide into 3 sections side 3 equals 0 In 2 EA 2EA From gauss law I q so oA so 2EA oA sol E o 2so Two parallel conducting infinite planes I EA q so 00 so 0 Spherical shell of charge q and radius R a Gaussian surface S which is a sphere with radius rgtR lnside equals 0 4rr2E CIEo E q4nr2so Uniformly charged sphere of radius R and charge q Outside E q4nr2so lnside 4nr2E qso qenc 4 qr34HR3 R3r3lq I 4nr2E R3r3qso Ei q4nsoR3r Electric potential The potential energy per unit charge The electric potential at a given point is the electric potential energy of a small test charge divided by the charge itself V EPEq units V The change in potential energy AU associated with a conservative force as the negative value of work W the force must do on a particle to take it from initial xi to nal position xf A positive charge accelerates from a region of higher electric potential toward a region of lower electric potential A negative charge accelerates from a region of lower potential toward a region of higher potential Equipotential surfaces A collection of points that have the same potential is known as an equipotential surface The work done by E as it moves a charge q between two points that have a potential difference AV is given by W q AV When a charge is moved on an equipotential surface AVO the work done by the electric eld is zero The electric eld E is perpendicular to the equipotential surfaces A charge q is moved by an electric eld E from point A to point B along a path Ar The electric field E forms an angle 6 with the path Ar W F Ar F Ar cose qE Ar cose Equals zero must be cose so 690 E q4nr220 Vp q4nao 1r R to infinity 14nao qR A conductor is an equipotential surface


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