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# FundamentalsofPhysicsII PHYS102

Drexel

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This 9 page Class Notes was uploaded by Vernice Schuster on Wednesday September 23, 2015. The Class Notes belongs to PHYS102 at Drexel University taught by SomTyagi in Fall. Since its upload, it has received 11 views. For similar materials see /class/212522/phys102-drexel-university in Physics 2 at Drexel University.

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Date Created: 09/23/15

Chapters23 and 24 You are expected to 1 Use Coulomb s law to calculate the electric forces between charges 2 Calculate the E eld due to a discrete charge distribution and simple continuous charge distributions 3 Know the distinction between the electric force F see eqsl 2 below and the E eld 4 Use Gauss law to calculate the E eld for symmetric continuous charge distributions 5 Solve problems that contain concepts from Chs23 and 24 some simple concepts from PHYS101 1 Coulomb s Law According to Coulomb s Law the magnitude of the electrostatic force between two charged particles with charges Q1 and Q2 and separated by a distance r is given by EkQ1 ZQ2 1 I Where k 9x1 09 N m2C2 is the Coulomb constant The unit of charge is taken as a Coulomb C The constant k is also written as k 14723980 where the constant 80 is called the permittivity offree space so 8854XlO3912 C2 Nm2 Since force is a vector quantity in the vector form Coulomb s law is expressed as I712 91 2 2 I I r F21 Q1 r12 Q2 F12 Where f is a unit vector directed from Q1 to Q2 The electric charge is quantized in units of l6103919 C magnitude ofthe charge of an electron Problem Solving 11 Finding force on a given charge due to a discrete charge distribution Break up the problem in three parts 1 Directiondraw the force vectors at a given charge location due to the other given charges Remember like charges repel unlike charges attract A welllabeled diagram will be very helpful 2Magnitude nd the magnitude of various force vectors using eql 3 a er step2 treat the problem as a vector manipulation problem The net force F on a given charge due to a discrete charge distribution is simply the vector sum of the forces produced by individual charges in the charge distribution at the location of the charge of your interest Study problem 2 below to clarify the points mentioned above 2 The E field The magnitude of the E eld generated by a charge Q is given by E kQI Z The direction of E at a given location is the direction in which a force would be exerted on a unit positive test charge placed at that location The E eld due to a discrete charge distribution is simply the vector sum of the E elds produced by individual charges in the charge distribution 21 Finding the E eld at a given location due to a discrete charge distribution Break up the problem in three parts 1 Direction draw the E eld vector directions at a given location due to the individual charges in the distribution Remember the direction of E at a given location is the direction in which a force would be exerted on a unit positive test charge placed at that location 2Magnitude nd the magnitude of various E eld vectors using E kQrz 3 after step 2 treat the problem as a vector manipulation problem Study problem 4 below to clarify the points mentioned above Problems 1 and 3 are problems that use concepts from Ch 19 and the conditions of translational equilibrium 2F ma 0 we studied last term You will need to brush up on your freebody force diagram FBD drawing abilities 3 Gauss Law 9 m GS 7 Gaussian surface 80 E GS In essence the total uX through a GS is determined solely by the charge enclosed by the GS A crucial aspect of applying Gauss law to determine the E eld due to a charge distribution is nding the charge enclosed Within a Gaussian surface Please DO NOT UNDERESTTMATE TPHS Here are a few cases to illustrate this point 1 A solid conducting sphere of radius R with charge Q Remember all charge will be forced to the surface of the sphere Why q 0 for r lt R 11m Q for r gt R 2 A solid dielectric insulator sphere of radius R with Q distributed unifome throughout the volume i rst determine the volume charge density p in terms of Q and R 0 3Q 4718 ii determine the charge enclosed by the Gaussian surface q 4727330 Qr3R3 for r lt R 11m Q for r gt R 3 A dielectric insulator spherical shell of inner radius a and outer radius 17 with Q distributed uniformly throughout the volume The volume charge density 0 3Q 47zb3 a3 Q q 0 for r lt a b qm 4m3 air3 Qr3 a3b3 a3 for alt r lt b q Q for r gt b How this information is used with Gauss law to find the E eld let s determine the E eld at various rvalues Region 1 r lt a IE1d Eljds qm 0 Note both E1 and ds point radially outward and therefore E1ds 80 Elds and since E1 is constant over the spherical GS E1 can be taken out of the integral E1 0 Region 2 alt r lt b 5265 Ezjds 47239r2E2 gm 80 where q 47Ir3 a303 Qr3 a3b3 a3 Therefore E2 14727280Qr3 a3b3 a3 kQr2r3 a3b3 a3 Region 3 r gt b E kQrz SOLVED EXAMPLES 1 Two identical cork balls each of mass m are hung from a common point by two insulating threads of negligible mass each oflength L 50 cm Each ball has a charge Q 02uC distributed unifome over its volume The balls repel each other and assume an equilibrium position as shown Determine the tension T in the threads and the mass m m Tsm60010 25x 10quot kg 39 T 39 60 i g 10m52 7 sm Solution 39 T Fcos600 Tcas60 0 2X9xj 09x4x103914052 39 3 300 300 F 288x10 N L l mg 10quot From T317160quot 10m r 2Lcas60 500m 2 Three charges are placed at the three vertices of a rightangle triangle as shown a Determine the force exerted on Q3 by charge Q1 Label this force F13 and eXpress it in the usual i j notation and show it on the diagram Y b Determine the force exerted on Q3 by charge Q2 Label this force F23 and express it in the usual i j notation and show it on the diagram c Determine the magnitude and direction of the net force on Q3 Solution Symbols in bold type are vectors a F13 9x1 098x625x1039125cos53 sz 2139 N 5001 N b F23 9x1094x625x10125xm 22 90N F23 90cos53i 7 90sm53 j 54p 72 jN c FT F23 F13 446p 72jN FT 446272212 452N 6 tanquot72446 92 3 A charge Q 3 0x10 8C is uniformly distributed throughout the volume of a small nonconducting ball of mass m 40 milligram The ball hangs as shown from an insulating thread of negligible mass in a region of uniform electrostatic E eld produced by a unifome charged plate ab In the diagram 6 370 and E eld is perpendicular to g and normal to the charged plate a In the space below draw a clearly labeled freebody force diagram for the ball b Determine the tension T in the thread and the magnitude of the E eld at the site of the ball Tcos37 T 7 Solution Tcos37 mg 4x10395N or T 4x10 5N cos37 5x105 N T5n37 qE Tsin37 qE or E Tsin37q 5x105x063x108 103NC 4X10395N 4 Three charges are placed on the three corners of a square as shown a Determine the E elds produced at the origin 0 individually by ql q and q3 E1 910930103960122j NC 1875106j NC E2 91098010396 20122 071 i ij NC 177 105 i j NC E3 91096010396012zi NC 375106i NC b Determine the net E eld produced at the origin by the three charges E E1 E2 E3 198 i0105j 106 NC c Determine the magnitude of the net E eld and its orientation with respect to the X aXlS E 1982 01052 2 106 NC 198 106 NC 6 tan3910105198 3 2 d If a charge Q 25uC is now placed at 0 What would be the net force on the charge F QE 495i 026 j N q2 80 c q1 30m q60 C 0 1 H 45 X i0 For Gauss Law problems 1 Study solved Examples 2435 of your Text 2 Review the case of a conducting sphere discussed in the class 3 Review the case of a nonconducting sphere discussed in the class 4 Study the case of a charged dielectric insulator spherical shell discussed above CONCEPTUAL QUESTIONS 1 Two positive point charges Q and 4Q are arranged as shown An electron is released from a point 10 cm away from charge Q Immediately after being released the E electron would move Q g 4Q A quot v g X a to the left lt2 i 4390 cm b to the right c the electron will be at equilibrium and stay where it is released 2 A thin uniformly charged ring lies in the x y plane with its center at the origin as shown The net E eld at point P on the zaxis will point quot P a somewhere in the x y plane b along the positive z axis axis c along the negative z axis axis d Efield at P is zero because of the symmetry of the charge distribution 3 An electric dipole is situated along the XaXis with its center at the origin The net E eld at point P on the yaXis will point a along the positive yaXis b along the positive XaXis c along the negative XaXis b The E eld at P is zero due to the opposite but equal contributions from the two charges 4 Two point charges are situated symmetrically about the origin The E eld measured at the origin is zero At point P the E eld points in the positive Xdirection From this one can conclude that a both charges are positive and equal in magnitude b both charges are negative and equal in magnitude c the charges are of opposite polarity but of equal magnitude d none of the above we don t have suf cient information 5 The E eld inside a uniformly charged conducting shell is determined to be zero When a positive charge Q is brought near the shell the Efield at the center of the shell would a still be zero b be non zero and point radially away from Q c be non zero and point radially toward Q

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