Block02Notes.pdf PHYS 242
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This 2 page Class Notes was uploaded by DB on Tuesday September 27, 2016. The Class Notes belongs to PHYS 242 at North Carolina A&T State University taught by Prof. Sandin in Fall 2016. Since its upload, it has received 7 views. For similar materials see General Physics II in Physics 2 at North Carolina A&T State University.
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Date Created: 09/27/16
PHYS 242 BLOCK 2 NOTES Sections 23.1 to 23.5 → Rewriting GeneralPhysics I’s Eq. (6.14) as Eq. (23.1), the wora→b done by a force F in moving a b ⌠ b particle from pointa to pointb is defined toa→bW≡ F·dl = ⌡F cos φ dl . a a If the force is conservatia→bW= U a U ,bwhereU is the potential energy associated with that force. b ⌠ If we have a test chargeq outside a spherically-symmetric, same-sign chargeq,= F·dl = 0 a→b a b 1 qq0 ⌡F cos φ dl , whereF =4πε0 r2 and dl cos φ = dr (see Fig. 23.6).Then integrating aromb to r gives us a 1 qq 0 1 qq 0 1 qq 0 4πε0 ra – 4πε0r b = Ua– U b Thus the simplest choice for the potential energy expression 4πε 0 r . This equation also holds for opposite-sign charges. Any equation in this block cε0is for vacuum ≈ air. U is the electric potential energy (in J) (TERM 1). 2 2 1 9 N·m 9 N·m Recall tha4πε0 = 8.988 × 10 C2 ≈ 9.0 × 10 C2 . q and q0are the charges (in C)—(point or spherically symmetric) (recall that charges can be +, 0, or –). r is the center-to-center distance (in m) between the two charges (recall thartis always+). Thus we see thatU is positive for same-sign charges and negative for opposite-sign charges. We also see thatU approaches zero as the distancer approaches infinity. This U can be used inKa+ U a K b U (Wb other 0) (SKILL 1), as in Example 23.1. The (electric) potential V at a point is the electric potential energy per charge at that point, V ≡ U (TERM 2) . Therefore, dividing both sides ofW = U – U by q tells us that the work percharge a→b is q0 a→b a b 0 q0 W a→b the (electric) potential differenceaV –bV between points a and b (TERM 3): = Va – b . We often q0 shorten V – V to V (and for electric circuits, often even furthert). jusV a b ab V is the (electric) potential (in V = volt = = J ) (V can be +, 0, or –). Please don’t mix upU, V, and V. coulomb C W a→b is the work done (in J) by the external electric field q 0rom a to b (W a→b can be +, 0, or –). Note that the two bold-face statements on page 759 (next to Fig. 23.11) are wrong b≠ J (units). 1 qq0 For a single point chargeq (or outside a spherically symmetric chargeq) we can substit4πε0U r to find V = = 1 q . For a collection of these charges, V = = Σ qi = 1 (q1 + q2 + …) . q0 4πε0r q 0 4πε0 ri 4πε 0 r1 r2 qiis the charge (in C) of objecii (q is +, 0, or –). riis the distance (in m) from the center of obiecti (r is always +). 1 dq For a continuous charge distribution, 4πε0 ∫ r . That is, first find an expression for the dV from an arbitrarydq in the distribution, then integrate to findV (SKILL 2). b b ⌡ → → ⌡ → → F·dl q0E·dl → Wa→b a a Finding a – b from E : Recall taatVb– Vq0 = q0 = q0 . Canceling ou0 q , we have → an equation we can use to find the potentiaif we know E at all points alongany path froma to b. b ⌠ → → b Va – b = E·dl = ⌡E cos φ dl . a a In this introductory course, we’ll replacedl withdx or dy or dz or dr in the integral. Note that E can have units of eiorer: 1 = 1J/C = 1N·m/ /= 1 N . m C m m / C Suppose that an electric field moves an electro–= –1.602 × 1019 C) through a potentialrise of one volt (boV is larger ahan V by 1 V). Then W a→b = q0(a – b ) = –e(–1 V) = 1 eV = (1.602 ×/ )(1 / ) = 1.602 × 10J. –19 That is, oneelectron volt(1 eV) is defined to equal 1J of work or energy (TERM 4). Cover up the solutions and carefully work Examples 23.3 to 23.11. The term equipotential means constant potential: V is constant at all points along an equipotential line, over an equipotential surface, or throughout an equipotential volume (TERMS 5, 6, and 7). For equipotential b ⌠ lines or surfacas,Vb– V = V – V = 0 = ⌡E cos φ dl tells us that φ must be 90˚ (if E ≠ 0). Thus, electric field a lines are always perpendicular to equipotential lines or equipotential surfaces. If its free charges are at rest overall, a conductingsurface is always an equipotentialsurface with any electricfield at its surface normal to → → that surface. Since the derivative of a constant is zero, E = – ∇ V (see below) tells us throughout an equipotential volume, E = 0. The electric field from a charged conducting surface tends to be greatest where the radius of curvature is smallest (for sharp points and thin wires). This large electric field can break down the air, giving an electrical discharge. In contrast, flatter areas tend to have smaller surface electric fields (SKILL 3). a b b → ⌠ ⌠ ⌡ → → → → Finding E from V: a – b = ⌡dV = (–dV) and V –aV =b E·dl tells us that –dV = E ·dl . b a a Evaluating the scalar (dot) product in terms of components (see Eq. (1.19)) gives –dV = E dx + E dy + E dz. x y z Keeping both y and z constant gives dy = 0 and dz = 0)y,z constantx. Such a derivativeis called a dx ∂V ∂V ∂V ∂V partial derivative. Thus we xave:∂x , Ey = –∂y , z = ∂z , and r = ∂r . Exis thex-component of the electric fielor () (with the same idea fory, z, or r) . m C Then E = E i + E j + E k = –( i ∂ + j ∂ + k ∂ )V. The quantity in the parentheses is the x y z ∂x ∂y ∂z gradient operator ∇(in Cartesian coordinates) , soE = – ∇ V. In words, the electricfield equals the negativeof the potential gradient (TERM 8). Cover up the solutions and carefully work Examples 23.13 and 23.14. Corrror 23.2 on page 784 by deleting “unit” twice and adding “per charge” after both “work” and “force”.
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