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Lecture 2 Electric Fields Chp. 22 Ed. 7

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Lecture 2 Electric Fields Chp. 22 Ed. 7 Lecture 2 Electric Fields Chp. 22 Ed. 7 • Cartoon - Analogous to gravitational field • Warm-up problems , Physlet • Topics – Electric field = Force per unit Charge – Electric Field Lines – Electric field from more than 1 charge – Electric Dipoles – Motion of point charges in an electric field – Examples of finding electric fields from continuous charges • List of Demos – Van de Graaff Generator, workings,lightning rod, electroscope, – Field lines using felt,oil, and 10 KV supply., • One point charge • Two same sign point charges • Two opposite point charges • Slab of charge – Smoke remover or electrostatic precipitator – Kelvin water drop generator – Electrophorus 1 2 Concept of the Electric Field • Definition of the electric field. Whenever charges are present and if I bring up another charge, it will feel a net Coulomb force from all the others. It is convenient to say that there is field there equal to the force per unit positive charge. E=F/q0. • The question is how does charge q0 know about 1charge q1 if it does not “touch it”? Even in a vacuum! We say there is a field produced by q1 that extends out in space everywhere. • The direction of the electric field is along r and points in the direction a positive test charge would move. This idea was proposed by Michael Faraday in the 1830’s. The idea of the field replaces the charges as defining the situation. Consider two point charges: r q1 q0 3 r q1 + q0 2 The Coulomb force is F= kq1q0/r The force per unit charge is E = F/q0 2 Then the electric field at r is E = kq1/r due to the point charge q1 . The units are Newton/Coulomb. The electric field has direction and is a vector. How do we find the direction.? The direction is the direction a unit positive test charge would move. This is called a field line. r E If q1 were positive q1 4 Example of field lines for a point negative charge. Place a unit q1 positive test chare at every point r and draw the direction r that it would move The blue lines are the field lines. The magnitude of the electric field is 2 E= kq1/r The direction of the field is given by the q1 line itself Important F= Eq0 , then ma=q0E, and then a = q0E/m 5 Electric Field Lines Like charges (++) Opposite charges (+ -) This is called an electric dipole. 6 Electric Field Lines: a graphic concept used to draw pictures as an aid to develop intuition about its behavior. The text shows a few examples. Here are the drawing rules. • E-field lines begin on + charges and end on - charges. (or infinity). • They enter or leave charge symmetrically. • The number of lines entering or leaving a charge is proportional to the charge • The density of lines indicates the strength of E at that point. • At large distances from a system of charges, the lines become isotropic and radial as from a single point charge equal to the net charge of the system. • No two field lines can cross. 7 Example of field lines for a uniform distribution of positive charge on one side of a very large nonconducting sheet. This is called a uniform electric field. 8 In order to get a better idea of field lines try this Physlet. • http://webphysics.davidson.edu/Applets/Applets.html • Click on problems • Click on Ch 9: E/M • Play with Physlet 9.1.4, 9.1.7 Demo: Show field lines using felt, oil, and 10 KV supply • One point charge • Two point charges of same sign • Two point charges opposite sign • Wall of charge 9 Methods of evaluating electric fields • Direct evaluation from Coulombs Law for a single point charge kq E = 1 r 2 1 • For a group of point charges, perform the vector sum N kq E = i ! r 2 i =1 i • This is a vector equation and can be complex and messy to evaluate and we may have to resort to a computer. The principle of superposition guarantees the result. 10 Typical Electric Fields (SI Units) 1 cm away from 1 nC of negative charge r . P - q E ' 10 Nm2 $ -9 %10 2 " !10 C kq C 5 N E = 1 = & # = 10 r 2 10-4 m2 C 11 Typical Electric Fields N Fair weather atmospheric electricity = 100 C downward 100 km high in the ionosphere + + + + + + + + + + + + + + + + + + + E Earth 12 Typical Electric Fields Field due to a proton at the location of the electron in the H atom. The radius of the electron orbit is 0.5 "10!10 m. ( 10 Nm2 % -19 &10 2 # "1.6 !10 C kq C 11 N E = 1 = ' $ = 4 !10 r 2 (0.5 !10-10 m)2 C - r Hydrogen atom + N Volt Note : = C meter 13 Example of finding electric field from two charges lying in a plane We have q 10 nC at the origin, q 15nC at x 4m 1 = + 2 = + = E x 0 What is at y = 3 m and = (at point P)? P q1 = +10nC q2 = +15nC Use principle of superposition Find x and y components of electric field due to both charges and add them up 14 E1 kq Example continued Recall E = r 2 2 E2 and k = 8.99 ! 109 Nm ( C2 ) Field due to q1 10 2 %'10 Nm "$ !10 !10(9 C C2 N ! E = & # = 11 in the y direction 1 (3m)2 C N E = 11 1y C ! E1x = 0 q =15 nc q1=10 nc 2 Field due to q2 E2 10 2 10 Nm ! 15 ! 10"9 C C 2 N E = ( ) = 6 at some angle # 2 (5m)2 C Now add all components Resolving it in x and y components : N N N 3 18 N E = (11+ 3.6) = 14.6 E = Esin# = 6 " = = 3.6 ynet 2y C 5 5 C C C N N ! 4 ! 24 N E = !4.8 E = Ecos# = 6 " = = !4.8 xnet 2x C 5 5 C C 15 N E E = 14.6 Example cont net y C !1 Magnitude of total electric field is N 2 2 Ex = !4.8 E = E x + E y C 2 2 E = (14.6) + (!4.8) =15.4N /C ! Direction of the total electric field is q =10 nc 1 q2 =15 nc "1 # Ey & "1 # 14.6 & !1 = tan = tan % ( = 107.2 or72.8° (degrees) $% E '( $ "4.8' x y Using unit vector notation we can also write the electric field vector as: ! ! " " j E = !4.8i+14.6 j ! i 16 Example of two identical Example of two opposite charges on the x axis. What charges on the x axis. What is the field on the y axis at P? is the field on the y axis at P? kq y y Ey E = 2 r P #x P 5 3 5 3 ! ! 4 4 x 4 4 x q = -15 nc q = +15 nc q2 = +15 nc q2 =+15 nc 2 2 E = 1010 N.m2/C2 15 X10-9 C/(5m)2 = 6 N/C Ey=2*E sin ! = 2"6 " 3/5 = 7.2 N/C #y=0 Ex=2*E cos ! = 2"6 " 4/5 = - 9.6 N/C #x=0 17 4 equal charges symmetrically spaced along a line. What is the field at point P? (y and x = 0) y P $4 r r1 4 $3 r $2 r3 $ 2 1 ! x q q3 q 1 q2 4 4 " 2 2 E y = k!qi cos"i /ri E y = k#!qi cos$i /ri i=1 i=1 18 Continuous distribution of charges • Instead of summing the charge we can imagine a continuous distribution and integrate it. This distribution may be over a volume, a surface or just a line. E = dE = kdq / r 2 ! ! dq = !dV volume dq = "dA area dq = #dl line 19 Find electric field due to a line of uniform + charge of length L with linear charge density equal to % y 2 dE = k dq /r L / 2 dE E = dE = 0 dE x " x y dEy= dE cos $ !L / 2 dEx r $0 y $ -x +x -L/2 0 x L/2 dq = %dx dq dE = k dx cos r2 2 y % $ / Ey= k % q cos $ /r for a point charge 2 L / 2 2 x= y tan$ dx = y sec $ d$ E y = k! $ cos" dx /r #L / 2 r =y sec $ r2 =y2sec2 $ 2 dx/r = d$/y " E k / y 0 cos d 2k! L /2 y = ! $ " " E = sin" sin!0 = #" 0 y 0 y 2 L2 /4 y + 20 What is the electric field from an infinitely long wire with linear charge density of +100 nC/m at a point 10 away from it. What do the lines of flux look like? ++++++++++++++++++++++++++++++++ $0 . y =10 cm &0=90 for an 2k! infinitely long wire E y = sin"0 y Ey 2*1010 Nm2 *100*10!9C /m E = sin90 = 2*104 N /C y 0.1m Typical field for the electrostatic smoke remover 21 Example of two opposite charges on the x axis. What is the field on the y axis? Electric field at point P at a distance r due to the two point charges L distance across is the sum of the electric fields due to +q and –q (superposition principle). Electric field due to the +q is E1 E1y k(+q) E = 1 a 2 1 E2x E1x Electric field due to the -q is E2 k(!q) E E 2y a 2 = 2 1 a2 a2 2 r Now, 2 & L # a1 = r + $ ! = a2 = a % 2 " Notice that the magnitude of E1 and E2 are equal.
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