Electrostatics: Coulomb's Law and Gauss' Law, Exams of Physics

Download Electrostatics: Coulomb's Law and Gauss' Law and more Exams Physics in PDF only on Docsity! Chapter-19 Coulomb’s Law You are expected to: 1. Use Coulomb’s law to calculate the electric forces between charges. 2. Calculate the E-field due to a discrete charge distribution and simple continuous charge distributions. 3. Know the distinction between the electric force F (see eqs.[1], [2] below) and the E-field 4. Use Gauss’ law to calculate the E-field for symmetric continuous charge distributions. 5. Solve problems that contain concepts from this chapter + some simple concepts from PHYS-101. 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: 1 22e Q QF k r = …[1] Where k = 9x109 N.m2/C2 is the Coulomb constant. The unit of charge is taken as a Coulomb (C). The constant k is also written as k = 1/4πεo where the constant εo is called the permittivity of free space εo = 8.854x10-12 C2/ N.m2. Since force is a vector quantity, in the vector form Coulomb’s law is expressed as: 1 212 122 ˆ Q QF k r = r …[2] r ++ r̂12Q1 Q2F21 F12 Where is a unit vector directed from Q12r̂ 1 to Q2. The electric charge is quantized in units of 1.6*10-19 C ( magnitude of the charge of an electron). Problem Solving: 1.1 Finding force on a given charge due to a discrete charge distribution. Break up the problem in three parts: [1] Direction:draw the force vectors at a given charge location due to the other given charges. (Remember: like charges repel, unlike charges attract.) A well-labeled diagram will be very helpful. [2]Magnitude: find the magnitude of various force vectors using eq.[1] [3] after step[2] 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-field generated by a charge Q is given by E = kQ/r2. 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-field due to a discrete charge distribution is simply the vector sum of the E-fields produced by individual charges in the charge distribution. 2.1 Finding the E-field at a given location due to a discrete charge distribution. Break up the problem in three parts: [1] Direction: draw the E-field 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). [2]Magnitude: find the magnitude of various E-field vectors using E = kQ/r2. [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 ( ΣF = ma = 0) we studied last term. You will need to brush up on your free-body- force diagram (FBD) drawing abilities. Chapter-19 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 of length L = 50 cm. Each ball has a charge Q = - 0.2μC distributed uniformly 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. 300 300 L m m g = 10m/s2 r = 2Lcos60 = 50cm mg =10m 60ο Τ Tsin60 Tcos60 F Solution: T = F/cos60o = 2x9x109x4x10-14/(0.5)2 = 2.88x10-3 N From Tsin60o = 10m, m = Tsin60o/10 = 2.5x 10-4 kg. 2. Three charges are placed at the three vertices of a right-angle 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. 53o 5.0 cm Y X Q 3= 6. 25 C Q2= 4.0 μC Q1= - 8.0 μC F13 F23 θ X 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 = - 9x109(8x6.25x10-12)/(5cos53 )2i N = - 500i N [b] F23 = 9x109(4x6.25x10-12)/(5x )2 = 90 N F23 = 90cos53i – 90sin53 j = 54i – 72 j N [c] FT = F23 +F13= - 446 i – 72 j N FT = {(446)2+(72)2}1/2 = 452 N , θ = tan-1(72/446) = 9.2o 3. A charge Q = 3.0x10-8C is uniformly distributed throughout the volume of a small non-conducting ball of mass, m = 4.0 milligram. The ball hangs, as shown, from an insulating thread of negligible mass in a region of uniform electrostatic E- field produced by a uniformly charged plate, ab. In the diagram, + + + + + + + E θ mQ a g =10m/s2 b θ = 37o, and E-field is perpendicular to g and normal to the charged plate. [a] In the space below, draw a clearly labeled free-body force diagram for the ball. [b] Determine the tension, T in the thread and the magnitude of the E-field at the site of the ball. T Tcos37 37 Tsin37 4x10-5N qE Solution: Tcos37 = mg = 4x10-5N or T = 4x10-5N /cos37 = 5x10-5 N Tsin37 = qE or E = Tsin37/q = 5x10-5x0.6/ 3x10-8 = 103N/C 4. Three charges are placed on the three corners of a square as shown. [a] Determine the E-fields produced at the origin, O individually by q1, q2 and q3. E1 = 9*109*3.0*10-6/(0.12)2j N/C = 1.875*106 j N/C E2 = [9*109*8.0*10-6/ 2(0.12)2 ]*0.71( - i – j) N/C = - 1.77 *106( i + j) N/C E3 = 9*109*6.0*10-6/(0.12)2i N/C =3.75*106 i N/C [b] Determine the net E-field produced at the origin by the three charges. E = E1+ E2+ E3 = (1.98 i +0.105 j ) 106 N/C [c] Determine the magnitude of the net E-field, and its orientation with respect to the x- axis. E = [ (1.98)2 + (0.105)2]1/2 * 106 N/C = 1.98 * 106 N/C θ = tan-1[0.105/1.98] =3. 2 o [d] If a charge Q = 2.5μC is now placed at O, what would be the net force on the charge? F = QE = ( 4.95i +0.26 j) N 450 E3 E2 E1 q1= - 3.0μC q2= 8.0μC q1= - 6.0μC y x12.0cm O