Work and Energy. v f. v i. F(x ) F(x ) x f. Consider a block of mass m moving along the x-axis. Conservative force acting on block: F = F(x) Z xf
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1 Work and Energy Consider a block of mass m moving along the x-axis. Conservative force acting on block: F = F(x) Work done by F(x) on block: W if = Kinetic energy of block: K = 1 2 mv2 Potential energy of block: U(x) = Z xf x i Z x F(x)dx x 0 F(x)dx F(x) = du dx Transformation of energy: K K f K i, U U f U i Total mechanical energy: E = K + U = const K + U = 0 Work-energy relation: W if = K = U v i v f m F(x ) i m F(x ) f x x i x f 2/9/2015 [tsl67 1/21]
2 Conservative Forces in Mechanics Conservative forces familiar from mechanics: Elastic force: F(x) = kx U(x) = Gravitational force (locally): F(y) = mg U(y) = Z y Z x x 0 ( kx)dx = 1 2 kx2 (x 0 = 0). y 0 ( mg)dy = mgy (y 0 = 0). Gravitational force (globally): F(r) = G mm E r 2 U(r) = Z r r 0 G mm E r 2 dr = G mm E r (r 0 = ). Potential energy depends on integration constant. Integration constant determines reference position where U = 0: x = x 0, y = y 0, r = r 0. 2/9/2015 [tsl68 2/21]
3 Work and Potential Energy in 3D Space Consider a particle acted on by a force F as it moves along a specific path in 3D space. Force: F = F x î + F y ĵ + F zˆk Displacement: d s = dxî + dyĵ + dzˆk Work: W if = Z rf Potential energy: U( r) = Z xf Z yf F d s = F x dx + F y dy + r i x i y i Z zf z i Z r Z x Z y F z dz F d s = F x dx F y dy r 0 x 0 y 0 z Z z z 0 F z dz Note: The work done by a conservative force is pathindependent. f F ds y x i 2/9/2015 [tsl69 3/21]
4 Potential Energy of Charged Particle in Uniform Electric Field Electrostatic force: F = qeĵ (conservative) Displacement: d s = dxî + dyĵ Work: W if = Z f i F d s = Potential energy: U = Z y Electric potential: V (y) = Ey 0 Z yf y i ( qe)dy = qe(y f y i ) ( qe)dy = qey y y i E ds i q F = qe y f f x 2/9/2015 [tsl70 4/21]
5 Potential Energy of Charged Particle in Coulomb Field Electrostatic force: F = kqq r 2 ˆr Displacement: d s = d r + d s, (conservative) d r = drˆr Z f Z f ˆr d s Work: W if = F d s = kqq i i r 2 = kqq» = kqq 1 rf» 1 = kqq 1 r r i r f r i Potential energy: U = Z r Fdr = kqq Z r dr Z rf r i r 2 = k qq r dr r 2 Electric potential: V (r) = kq r dr q ds f i r r f. r i + Q 2/9/2015 [tsl71 5/21]
6 Attributes of Space and of Charged Particles planar source point source SI unit electric field E = Ex î E = kq r 2 ˆr electric potential V = E x x V = kq r electric force F = qe = qex î F = qe kqq = r 2 ˆr electric potential energy U = qv = qe x x U = qv = kqq r [N/C]=[V/m] [V]=[J/C] [N] [J] Electric field E is present at points in space. Points in space are at electric potential V. Charged particles experience electric force F = q E. Charged particles have electric potential energy U = qv. 2/9/2015 [tsl447 6/21]
7 Equipotential Surfaces and Field Lines Definition: V ( r) = const on equipotential surface. Potential energy U( r) = const for point charge q on equipotential surface. The surface of a conductor at equilibrium is an equipotential surface. Electric field vectors E( r) (tangents to field lines) are perpendicular to equipotential surface. Electrostatic force F = q E( r) does zero work on point charge q moving on equipotential surface. The electric field E( r) exerts a force on a positive (negative) point charge q in the direction of steepest potential drop (rise). When a positive (negative) point charge q moves from a region of high potential to a region of low potential, the electric field does positive (negative) work on it. In the process, the potential energy decreases (increases). 2/9/2015 [tsl80 7/21]
8 Topographic Maps Gravitation Electricity 2/9/2015 [tsl421 8/21]
9 Electric Potential and Potential Energy: Application (1) Consider a point charge Q = 2µC fixed at position x = 0. A particle with mass m = 2g and charge q = 0.1µC is launched at position x 1 = 10cm with velocity v 1 = 12m/s. (fixed) Q = 2µC x = 0 m = 2g q = 0.1µC v 1 x 1= 10cm x 2= 20cm Find the velocity v 2 of the particle when it is at position x 2 = 20cm. 2/9/2015 [tsl73 9/21]
10 Electric Potential and Potential Energy: Application (2) Electric potential at point P 1 : V = kq m + kq m Electric potential at point P 2 : V = kq kq m = 1125V V = 2250V. = 750V + 450V = 1200V. 2/9/2015 [tsl74 10/21]
11 Electric Potential and Potential Energy: Application (3) Point charges q 1 = 5.0µC and q 2 = +2.0µC are positioned at two corners of a rectangle as shown. q = 5.0µC 1 15cm 5cm A B q =+2.0µC 2 (a) Find the electric potential at the corners A and B. (b) Find the electric field at point B. (c) How much work is required to move a point charge q 3 = +3µC from B to A? 2/9/2015 [tsl75 11/21]
12 Electric Potential and Potential Energy: Application (4) A positive point charge q is positioned in the electric field of a negative point charge Q. (1) Q = 10µC 2m q = +5µC (2) Q = 10µC 1m q = +1µC (a) In which configuration is the charge q positioned in the stronger electric field? (b) In which configuration does the charge q experience the stronger force? (c) In which configuration is the charge q positioned at the higher electric potential? (d) In which configuration does the charge q have the higher potential energy? 2/9/2015 [tsl76 12/21]
13 Electric Potential and Potential Energy: Application (5) An electron and a proton are released from rest midway between oppositely charged plates E (a) Name the particle(s) which move(s) from high to low electric potential. (b) Name the particle(s) whose electric potential energy decrease(s). (c) Name the particle(s) which hit(s) the plate in the shortest time. (d) Name the particle(s) which reach(es) the highest kinetic energy before impact. 2/9/2015 [tsl77 13/21]
14 Electric Potential and Potential Energy: Application (6) Three protons are projected from x = 0 with equal initial speed v 0 in different directions. They all experience the force of a uniform horizontal electric field E. Ultimately, they all hit the vertical screen at x = L. p1 p2 E p3 x = 0 x = L x (a) Which proton travels the longest time? (b) Which proton travels the longest path? (c) Which particle has the highest speed when it hits the screen? Two of the questions are easy, one is hard. 2/9/2015 [tsl78 14/21]
15 Electric Potential and Potential Energy: Application (7) Consider a region of nonuniform electric field. Charged particles 1 and 2 start moving from rest at point A in opposite directions along the paths shown. C 2 q = 1C 2 A V = 17V A 1 q = +3C 1 B V = 11V B From the information given in the figure... (a) find the kinetic energy K 1 of particle 1 when it arrives at point B, (b) find the electric potential V C at point C if we know that particle 2 arrives there with kinetic energy K 2 = 8J. 2/9/2015 [tsl79 15/21]
16 Electric Potential and Potential Energy: Application (8) (a) Is the electric potential at points P 1, P 2 positive or negative or zero? (b) Is the potential energy of a negatively charged particle at points P 1, P 2 positive or negative or zero? (c) Is the electric field at points P 1, P 2 directed left or right or is it zero? (d) Is the force on a negatively charged particle at points P 1 and P 2 directed left or right or is it zero? 2nC P 1 4nC 3cm 6cm 2nC P 2 4nC 3cm 6cm 2/9/2015 [tsl83 16/21]
17 Electric Potential and Potential Energy: Application (9) Consider four point charges of equal magnitude positioned at the corners of a square as shown. Answer the following questions for points A, B, C. (1) Which point is at the highest electric potential? (2) Which point is at the lowest electric potential? (3) At which point is the electric field the strongest? (4) At which point is the electric field the weakest? +q A +q B q C q 2/9/2015 [tsl84 17/21]
18 Electric Potential and Potential Energy: Application (10) The charged particles 1 and 2 move between the charged conducting plates A and B in opposite directions. From the information given in the figure... (a) find the kinetic energy K 1B of particle 1, (b) find the charge q 2 of particle 2, (c) find the direction and magnitude of the electric field E between the plates. K1A =3µJ q 1 = 2µC 1 K 1B =? K 2A = 10µ J V = 19V A q 2 =? 2 2m V = 13V B K 2B = 4µ J 2/9/2015 [tsl90 18/21]
19 Intermediate Exam I: Problem #2 (Spring 05) Consider a point charge Q = 5nC fixed at position x = 0. (a) Find the electric potential V 1 at position x 1 = 3m and the electric potiential V 2 at position x 2 =. (b) If a charged particle (q = 4nC, m = 1.5ng) is released from rest at x 1, what are its kinetic energy K 2 and its velocity v 2 when it reaches position x 2? Q = 5nC x = 0 x 1= 3m x 2= 2/9/2015 [tsl332 19/21]
20 Intermediate Exam I: Problem #2 (Spring 05) Consider a point charge Q = 5nC fixed at position x = 0. (a) Find the electric potential V 1 at position x 1 = 3m and the electric potiential V 2 at position x 2 =. (b) If a charged particle (q = 4nC, m = 1.5ng) is released from rest at x 1, what are its kinetic energy K 2 and its velocity v 2 when it reaches position x 2? Q = 5nC Solution: x = 0 x 1= 3m x 2= (a) V 1 = k Q x 1 = 15V, V 2 = k Q x 2 = 7.5V. 2/9/2015 [tsl332 19/21]
21 Intermediate Exam I: Problem #2 (Spring 05) Consider a point charge Q = 5nC fixed at position x = 0. (a) Find the electric potential V 1 at position x 1 = 3m and the electric potiential V 2 at position x 2 =. (b) If a charged particle (q = 4nC, m = 1.5ng) is released from rest at x 1, what are its kinetic energy K 2 and its velocity v 2 when it reaches position x 2? Q = 5nC Solution: x = 0 x 1= 3m x 2= (a) V 1 = k Q x 1 = 15V, V 2 = k Q x 2 = 7.5V. (b) U = q(v 2 V 1 ) = (4nC)( 7.5V) = 30nJ K = U = 30nJ. K = K 2 = 1 r 2K2 2 mv2 2 v 2 = m = 200m/s. 2/9/2015 [tsl332 19/21]
22 Unit Exam I: Problem #1 (Fall 10) Consider two point charges positioned as shown. +7nC (a) Find the magnitude of the electric field at point A. (b) Find the electric potential at point A. (c) Find the magnitude of the electric field at point B. (d) Find the electric potential at point B. A B 7nC 2/9/2015 [tsl398 20/21]
23 Unit Exam I: Problem #1 (Fall 10) Consider two point charges positioned as shown. +7nC (a) Find the magnitude of the electric field at point A. (b) Find the electric potential at point A. (c) Find the magnitude of the electric field at point B. (d) Find the electric potential at point B. A Solution: (a) E A = 2k 7nC = 2(2.52V/m) = 5.04V/m. () 2 B 7nC 2/9/2015 [tsl398 20/21]
24 Unit Exam I: Problem #1 (Fall 10) Consider two point charges positioned as shown. +7nC (a) Find the magnitude of the electric field at point A. (b) Find the electric potential at point A. (c) Find the magnitude of the electric field at point B. (d) Find the electric potential at point B. A Solution: (a) E A = 2k 7nC = 2(2.52V/m) = 5.04V/m. () 2 B 7nC (b) V A = k (+7nC) + k ( 7nC) = 12.6V 12.6V = 0. 2/9/2015 [tsl398 20/21]
25 Unit Exam I: Problem #1 (Fall 10) Consider two point charges positioned as shown. (a) Find the magnitude of the electric field at point A. (b) Find the electric potential at point A. (c) Find the magnitude of the electric field at point B. (d) Find the electric potential at point B. Solution: (a) E A = 2k 7nC = 2(2.52V/m) = 5.04V/m. () 2 (b) V A = k (+7nC) + k ( 7nC) s (c) E B = k 7nC () 2 «2 + k 7nC () 2 = 12.6V 12.6V = 0. +7nC B A 7nC «2 E B = q (1.75V/m) 2 + (0.98V/m) 2 = 2.01V/m. 2/9/2015 [tsl398 20/21]
26 Unit Exam I: Problem #1 (Fall 10) Consider two point charges positioned as shown. (a) Find the magnitude of the electric field at point A. (b) Find the electric potential at point A. (c) Find the magnitude of the electric field at point B. (d) Find the electric potential at point B. Solution: (a) E A = 2k 7nC = 2(2.52V/m) = 5.04V/m. () 2 (b) V A = k (+7nC) + k ( 7nC) s (c) E B = k 7nC () 2 (d) V B = k (+7nC) «2 + k 7nC () 2 + k( 7nC) = 12.6V 12.6V = 0. +7nC B A 7nC «2 E B = q (1.75V/m) 2 + (0.98V/m) 2 = 2.01V/m. = 10.5V 7.9V = 2.6V. 2/9/2015 [tsl398 20/21]
27 Unit Exam I: Problem #1 (Spring 14) Consider two point charges positioned as shown. Find the magnitude of the electric field at point A. Find the electric potential at point B. Find the magnitude of the electric field at point C. Find the electric potential at point D. +5nC 3m A 3m C 4m D B 9nC 2/9/2015 [tsl469 21/21]
28 Unit Exam I: Problem #1 (Spring 14) Consider two point charges positioned as shown. Find the magnitude of the electric field at point A. Find the electric potential at point B. Find the magnitude of the electric field at point C. Find the electric potential at point D. +5nC 3m A 3m C 4m D Solution: B 9nC E A = k 5nC (3m) 2 + k 9nC (7m) 2 = 5.00V/m V/m = 6.65V/m. 2/9/2015 [tsl469 21/21]
29 Unit Exam I: Problem #1 (Spring 14) Consider two point charges positioned as shown. Find the magnitude of the electric field at point A. Find the electric potential at point B. Find the magnitude of the electric field at point C. Find the electric potential at point D. +5nC 3m A 3m C 4m D Solution: B 9nC E A = k 5nC (3m) 2 + k 9nC (7m) 2 V B = k (+5nC) + k( 9nC) = 5.00V/m V/m = 6.65V/m. = 7.50V 10.13V = 2.63V. 2/9/2015 [tsl469 21/21]
30 Unit Exam I: Problem #1 (Spring 14) Consider two point charges positioned as shown. Find the magnitude of the electric field at point A. Find the electric potential at point B. Find the magnitude of the electric field at point C. Find the electric potential at point D. +5nC 3m A 3m C 4m D B 9nC Solution: E A = k 5nC (3m) 2 + k 9nC (7m) 2 = 5.00V/m V/m = 6.65V/m. V B = k (+5nC) + k( 9nC) E C = k 5nC () 2 + k 9nC (4m) 2 = 7.50V 10.13V = 2.63V. = 1.25V/m V/m = 6.31V/m. 2/9/2015 [tsl469 21/21]
31 Unit Exam I: Problem #1 (Spring 14) Consider two point charges positioned as shown. Find the magnitude of the electric field at point A. Find the electric potential at point B. Find the magnitude of the electric field at point C. Find the electric potential at point D. +5nC 3m A 3m C 4m D B 9nC Solution: E A = k 5nC (3m) 2 + k 9nC (7m) 2 = 5.00V/m V/m = 6.65V/m. V B = k (+5nC) + k( 9nC) E C = k 5nC () 2 + k 9nC (4m) 2 = 7.50V 10.13V = 2.63V. = 1.25V/m V/m = 6.31V/m. V D = k (+5nC) + k ( 9nC) = 5.63V 13.5V = 7.87V. 2/9/2015 [tsl469 21/21]
05. Electric Potential I
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