AP Physics C: Mechanics:: Formula Sheet, Slides of Physics

Download AP Physics C: Mechanics:: Formula Sheet and more Slides Physics in PDF only on Docsity! AP Physics C: Mechanics:: Formula Sheet I II III IV V VI Uniform fields or nonlinear fields that are drawn. Point charge q superposition Sphere with charge Q and sphere within a sphere Arc with Point P at center of the arc Hoop with P offset along an axis passing thru center Cylinder E given (or V is given) If V is given instead of E you can use the equation E = − 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 to solve for E . Usually these problems involve uniform fields or when non-uniform they allow you to find average E between two equal potential line values. 𝐸𝐸 = 𝑑𝑑 𝑑𝑑 All four of these problems use the same equation derived from Gauss’s Law ∮𝐵𝐵�⃗ ∙ 𝑑𝑑𝐴𝐴�����⃗ = 𝑄𝑄𝑒𝑒𝑒𝑒𝑒𝑒 𝜖𝜖0 𝐸𝐸(4𝜋𝜋𝑑𝑑2) = 𝑄𝑄𝑒𝑒𝑒𝑒𝑒𝑒 𝜖𝜖0 𝐸𝐸 = 1 4𝜋𝜋𝜖𝜖0 𝑄𝑄𝑒𝑒𝑒𝑒𝑒𝑒 𝜖𝜖0 𝑘𝑘 = 1 4𝜋𝜋 𝜖𝜖0 �𝐵𝐵�⃗ ∙ 𝑑𝑑𝐴𝐴�����⃗ = 𝑄𝑄𝑒𝑒𝑒𝑒𝑒𝑒 𝜖𝜖0 𝐸𝐸(2𝜋𝜋𝑑𝑑𝜋𝜋) = 𝑄𝑄𝑒𝑒𝑒𝑒𝑒𝑒 𝜖𝜖0 𝐸𝐸 = 𝑄𝑄𝑒𝑒𝑒𝑒𝑒𝑒 2𝜋𝜋𝑑𝑑𝜋𝜋𝜖𝜖0 𝜆𝜆 = 𝑄𝑄 𝐿𝐿 𝐸𝐸 = 𝜆𝜆 2𝜋𝜋𝑑𝑑𝜖𝜖0 Use 𝐸𝐸 = 𝑘𝑘 𝑄𝑄 𝑟𝑟² to find All the individual 𝐸𝐸𝑖𝑖 , vectors due to each charge 𝑞𝑞𝑖𝑖 acting a point P. Add all the individual 𝐸𝐸𝑖𝑖 vectors to find the total electric field E. If a new charge appears at point P, then you can find the force on the new charge due to the field E . 𝐹𝐹𝐸𝐸 = 𝑞𝑞𝑒𝑒𝑒𝑒𝑛𝑛𝐸𝐸𝑡𝑡𝑡𝑡𝑡𝑡𝑎𝑎𝑙𝑙 Use 𝐸𝐸 = 𝑘𝑘 𝑄𝑄𝑒𝑒𝑒𝑒𝑒𝑒 𝑟𝑟²𝐺𝐺𝐺𝐺 to find E at various points (inside the sphere r < R , on its surface r = R, and outside the sphere r > R ). 𝐸𝐸 = 𝑘𝑘 𝑄𝑄 𝑑𝑑² 𝑑𝑑𝐸𝐸 = 𝑘𝑘 𝑑𝑑𝑞𝑞 𝑑𝑑² �𝑑𝑑𝐸𝐸𝑥𝑥 = �𝑘𝑘 𝑑𝑑𝑞𝑞 𝑑𝑑² cos 𝜃𝜃 Density 𝜆𝜆 = 𝑄𝑄 𝐿𝐿 = 𝑑𝑑𝑞𝑞 𝑑𝑑𝜋𝜋 𝐸𝐸𝑥𝑥 = � 𝑘𝑘𝑄𝑄 𝑑𝑑2𝐿𝐿 𝑑𝑑𝜋𝜋 cos 𝜃𝜃 𝑑𝑑 = 𝑅𝑅 Arc length 𝑑𝑑𝜋𝜋 = 𝑅𝑅𝑑𝑑𝜃𝜃 𝐸𝐸𝑥𝑥 = � 𝑘𝑘𝑄𝑄 𝑅𝑅𝐿𝐿 cos𝜃𝜃 𝑑𝑑𝜃𝜃 𝐸𝐸𝑥𝑥 = 𝑘𝑘𝑄𝑄 𝑅𝑅𝐿𝐿 � cos𝜃𝜃 𝑑𝑑𝜃𝜃 𝜃𝜃 𝜃𝜃0 𝐸𝐸𝑥𝑥 = 𝑘𝑘𝑄𝑄 𝑅𝑅𝐿𝐿 (sin𝜃𝜃𝑏𝑏 − sin𝜃𝜃𝑎𝑎) 𝐸𝐸 = 𝑘𝑘 𝑄𝑄 𝑑𝑑² 𝑑𝑑𝐸𝐸 = 𝑘𝑘 𝑑𝑑𝑞𝑞 𝑑𝑑² �𝑑𝑑𝐸𝐸𝑥𝑥 = �𝑘𝑘 𝑑𝑑𝑞𝑞 𝑑𝑑² cos 𝜃𝜃 𝐸𝐸𝑥𝑥 = 𝑘𝑘 𝑑𝑑² cos 𝜃𝜃�𝑑𝑑𝑞𝑞 𝐸𝐸𝑥𝑥 = 𝑘𝑘𝑄𝑄 𝑑𝑑² cos 𝜃𝜃 cos 𝜃𝜃 = 𝑥𝑥 𝑟𝑟 and 𝑑𝑑 = �𝑅𝑅2 + 𝑥𝑥² 𝐸𝐸𝑥𝑥 = 𝑘𝑘𝑄𝑄𝑥𝑥 (𝑅𝑅2 + 𝑥𝑥2)3/2 r < R in an insulator 𝑄𝑄𝑒𝑒𝑒𝑒𝑒𝑒 𝑑𝑑𝑒𝑒𝑒𝑒𝑒𝑒 = 𝑄𝑄𝑡𝑡𝑡𝑡𝑡𝑡𝑎𝑎𝑙𝑙 𝑑𝑑𝑡𝑡𝑡𝑡𝑡𝑡𝑎𝑎𝑙𝑙 𝑄𝑄𝑒𝑒𝑒𝑒𝑒𝑒 �4 3�𝜋𝜋𝑑𝑑³ = 𝑄𝑄𝑡𝑡𝑡𝑡𝑡𝑡𝑎𝑎𝑙𝑙 �4 3�𝜋𝜋𝑑𝑑³ r < R in an insulator 𝑄𝑄𝑒𝑒𝑒𝑒𝑒𝑒 𝑑𝑑𝑒𝑒𝑒𝑒𝑒𝑒 = 𝑄𝑄𝑡𝑡𝑡𝑡𝑡𝑡𝑎𝑎𝑙𝑙 𝑑𝑑𝑡𝑡𝑡𝑡𝑡𝑡𝑎𝑎𝑙𝑙 𝑄𝑄𝑒𝑒𝑒𝑒𝑒𝑒 𝜋𝜋𝑑𝑑²𝐿𝐿 = 𝑄𝑄𝑡𝑡𝑡𝑡𝑡𝑡𝑎𝑎𝑙𝑙 𝜋𝜋𝑑𝑑²𝐿𝐿 E = − 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑 = −�𝐸𝐸 𝑑𝑑𝑑𝑑 Usually these problems involve uniform fields. 𝑑𝑑 = 𝐸𝐸𝑑𝑑 𝑑𝑑 = 𝑘𝑘Σ 𝑞𝑞𝑖𝑖 𝑑𝑑𝑖𝑖 𝑑𝑑𝑡𝑡𝑡𝑡𝑡𝑡𝑎𝑎𝑙𝑙 = 𝑘𝑘 � 𝑞𝑞1 𝑑𝑑1 + 𝑞𝑞2 𝑑𝑑2 + 𝑞𝑞3 𝑑𝑑3 +. . � If a new charge appears at point P, then you can find the energy on the new charge due to the total potential 𝑑𝑑 . 𝑈𝑈𝐸𝐸 = 𝑞𝑞𝑒𝑒𝑒𝑒𝑛𝑛𝑑𝑑𝑡𝑡𝑡𝑡𝑡𝑡𝑎𝑎𝑙𝑙 If released the charge is released you can find its speed 𝑞𝑞𝑒𝑒𝑒𝑒𝑛𝑛𝑑𝑑𝑡𝑡𝑡𝑡𝑡𝑡𝑎𝑎𝑙𝑙 = 1 2 𝑚𝑚𝑒𝑒𝑒𝑒𝑛𝑛𝑣𝑣²𝑒𝑒𝑒𝑒𝑛𝑛 Change in potential moving from point a to b 𝛥𝛥𝑑𝑑 = −� 𝐸𝐸 𝑏𝑏 𝑎𝑎 𝑑𝑑𝑑𝑑 𝛥𝛥𝑑𝑑 = − � �𝑘𝑘 𝑄𝑄 𝑑𝑑2 � 𝑏𝑏 𝑎𝑎 𝑑𝑑𝑑𝑑 𝑑𝑑 = − 𝑘𝑘𝑄𝑄� � 1 𝑑𝑑2 � 𝑏𝑏 𝑎𝑎 𝑑𝑑𝑑𝑑 𝑑𝑑 = 𝑘𝑘 𝑄𝑄𝑒𝑒𝑒𝑒𝑒𝑒 𝑑𝑑 �𝑏𝑏𝑎𝑎 𝑑𝑑 = 𝑘𝑘𝑄𝑄 � 1 𝑏𝑏 − 1 𝑎𝑎 � (If a = ∞ and b = R ) 𝑑𝑑 = 𝑘𝑘 𝑄𝑄 𝑅𝑅 The arc consists of many charged spheres (protons or electrons) to total. 𝑑𝑑 = 𝑘𝑘 � 𝑞𝑞1 𝑑𝑑1 + 𝑞𝑞2 𝑑𝑑2 + 𝑞𝑞3 𝑑𝑑3 +. . � All points are at the same distance r , and R = r . 𝑑𝑑 = 𝑘𝑘 𝑄𝑄 𝑅𝑅 The hoop consists of many charged spheres (protons or electrons) to total. 𝑑𝑑 = 𝑘𝑘 � 𝑞𝑞1 𝑑𝑑1 + 𝑞𝑞2 𝑑𝑑2 + 𝑞𝑞3 𝑑𝑑3 +. . � 𝑑𝑑 = 𝑘𝑘 𝑄𝑄 𝑅𝑅 All points are at the same distance r , where 𝑑𝑑 = �𝑅𝑅2 + 𝑥𝑥2 𝑑𝑑 = 𝑘𝑘 𝑄𝑄 �𝑅𝑅2 + 𝑥𝑥² Change in potential moving from point a to b 𝑑𝑑 = −� 𝐸𝐸 𝑏𝑏 𝑎𝑎 𝑑𝑑𝑑𝑑 𝑑𝑑 = −� � 𝜆𝜆 2𝜋𝜋𝑑𝑑𝜖𝜖0 � 𝑏𝑏 𝑎𝑎 𝑑𝑑𝑑𝑑 𝑑𝑑 = − 𝜆𝜆 2𝜋𝜋𝑑𝑑𝜖𝜖0 � � 1 𝑑𝑑 �𝑑𝑑𝑑𝑑 𝑏𝑏 𝑎𝑎 𝑑𝑑 = − 𝜆𝜆 2𝜋𝜋𝜖𝜖0 (𝐼𝐼𝐼𝐼 𝑏𝑏 − 𝐼𝐼𝐼𝐼 𝑎𝑎 𝑑𝑑 = − 𝜆𝜆 2𝜋𝜋𝜖𝜖0 𝐼𝐼𝐼𝐼 𝑏𝑏 𝑎𝑎 2005-1 2000-2, 2001-1, 2006-1 1996-1, 1997-2, 1998-1, 1999-1, 2003-1, 2004-1, 2007-2, 2008-1 2002-1 1999-3 1998-1, 2000-3 I II III IV V VI Fixed Magnet Wires and Superposition Cylinder with current 𝐼𝐼 Cylinder within a cylinder Arc with Point P at center of the arc Hoop with P offset along an axis passing thru center Solenoid These two problems use Ampere’s Law �𝐵𝐵�⃗ ∙ 𝑑𝑑𝜋𝜋���⃗ = 𝜇𝜇0𝐼𝐼𝑒𝑒𝑒𝑒𝑒𝑒 𝐵𝐵�⃗ �𝑑𝑑𝜋𝜋���⃗ = 𝜇𝜇0𝐼𝐼𝑒𝑒𝑒𝑒𝑒𝑒 In Ampere’s Law 𝑑𝑑𝜋𝜋 is the length of the magnetic field which circles the wire, with a sum of ∮𝑑𝑑𝜋𝜋���⃗ = 2𝜋𝜋𝑑𝑑 Lower case 𝑑𝑑 is used since we are measuring the to These two problems use Biot-Savart 𝑑𝑑𝐵𝐵�����⃗ = 𝜇𝜇0 4𝜋𝜋 𝐼𝐼𝑑𝑑𝜋𝜋���⃗ × ?̂?𝑑 𝑑𝑑² 𝑑𝑑𝐵𝐵�����⃗ = 𝜇𝜇0 4𝜋𝜋 𝐼𝐼𝑑𝑑𝜋𝜋���⃗ 𝑑𝑑² sin𝜃𝜃 𝐵𝐵𝑠𝑠 = 𝜇𝜇0𝐼𝐼𝐼𝐼 𝐼𝐼 = 𝑁𝑁 𝐿𝐿