Download Charge Distribution - Electricity and Magnetism - Solved Past Paper and more Exams Electromagnetism and Electromagnetic Fields Theory in PDF only on Docsity! Physics 4B FIRST EXAM Chapters 23 - 25 Fall 1996 1 Name:_________________________________________ Solve the following problems in the space provided. Use the back of the page if needed. Each problem is worth 20 points. You must show your work in a logical fashion starting with the correctly applied physical principles which are on the last page. Your score will be maximized if your work is easy to follow because partial credit will be awarded. For full credit you must explain clearly what you are doing especially if your solution involves symmetry arguments or uses Gauss's Law. 1. The two charges q are each a distance a to the right and left of the origin as shown at the right. Find the magnitude and direction of the electric force on the charge -Q which is a very small distance y above the origin. Show that this force varies linearly with y. The magnitudes of the two forces can be found with Coulomb's Rule, r F e = k q1q2 2r ˆ r ! F1 = F2 = k qQ a2 + y2 These must be added as vectors, F x = F 2 cos!" F 1 cos ! = 0 as might be expected from symmetry considerations. Fy = !F2 sin" ! F1 sin" = !2k qQ a2 + y2 sin" Note that sin! = y a2 + y2 . The net force is downward with a magnitude of F = 2k qQy a 2 + y 2( ) 3 2 This is not linear in y, but for y very small compared to a, the y in the denominator can be neglected giving F = 2k qQy a3 which is linear in y. -Q q q aa yF 1 F 2 ! ! Physics 4B FIRST EXAM Chapters 23 - 25 Fall 1996 2 2. A solid metal sphere of radius 5.00cm creates an electric field of 2.25x106N/C at a distance of 20.0cm from its center. Describe the charge distribution on the sphere. Start by applying Gauss's Law, r E • d r A ! = q "o , to an imaginary sphere 20.0cm in radius centered on the conducting sphere. By the spherical symmetry of the problem, the field must be radial and constant on this imaginary sphere so, EA = q !o " q = !oEA = !oE4#r 2 q = 8.85x10 $12( ) 2.25x106( )4# 0.200( )2 = 10.0µC Using Gauss's Law and the same symmetry arguments for an imaginary sphere inside the metal sphere, EA = q !o " q = !oEA . However, the field inside a conductor is always zero, so the charge inside is also zero. Therefor, all the charge on the sphere is located on the surface. 3. The principle that the book calls "Coulomb's Law" I refer to as "Coulomb's Rule." I do this because laws cannot be derived while rules can be derived from laws and definitions. Coulomb’s Rule can be derived by starting with a point charge and applying Gauss's Law. Complete this derivation by explaining the steps carefully. The field due to q1 can be found by applying Gauss's Law, r E • d r A ! = q "o , to an imaginary sphere of radius r centered on the charge. By the spherical symmetry of the problem, the field must be radial and constant on this imaginary sphere so, E4!r2 = q1 "o # E = q1 4!" or 2 = k q1 r2 . Now suppose a charge q2 is in the field of q1 a distance r away. The force on the q2 can be found, by the definition of electric field. E ! F q " k q1 r2 = F q2 " F = k q1q2 r2 which is precisely Coulomb’s Rule. gaussian sphere conducting sphere gaussian sphere conducting sphere gaussian sphere point charge q1 r q1 q2 r
