# Electric Fields Bending of water stream with charged rod Warm-up problem

## Electric Fields Bending of water stream with charged rod Warm-up problem

Walsh, Natalie, Executive Producer has reference to this Academic Journal, PHwiki organized this Journal Lecture 2 Electric Fields Chp. 23 Cartoon – Analogous to gravitational field Opening Demo – Bending of water stream with charged rod Warm-up problem Physlet Topics Electric field = Force per unit Charge Electric Field Lines in addition to Electric Flux 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, electric wind Smoke remover or electrostatic precipitator Kelvin water drop generator Transparent CRT with visible electron gun Field lines using felt,oil, in addition to 10 KV supply. The Electric Field Definition of the electric field. Whenever charges are present in addition to if I bring up another charge, it will feel a net Coulomb as long as ce from all the others. It is convenient to say that there is field there equal to the as long as ce per unit positive charge. E=F/q0. The direction of the electric field is along r in addition to 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: The as long as ce per unit charge is E = F/q0 in addition to then the electric field at r is E = kq1/r2 due to the point charge q1 . The units are Newton/Coulomb. The electric field has direction in addition to is a vector. How do we find the direction. The direction is the direction a unit positive test charge would move. The Coulomb as long as ce is F= kq1q0/r2 If q1 were positive

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Point negative charge E= kq1/r2 q1 q1 r Electric Field Lines Like charges (++) Opposite charges (+ -) This is called an electric dipole. 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 in addition to 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 in addition to radial as from a single point charge equal to the net charge of the system. No two field lines can cross. Show a physlet 9.1.4, 9.1.7 Show field lines using felt,oil, in addition to 10 KV supply

Typical Electric Fields (SI Units) 1 cm away from 1 nC of negative charge E = kq /r2 = 1010 10-9/ 10-4 =105 N /C N.m2/C2 C / m2 = N/C Fair weather atmospheric electricity = 100 N/C downward 100 km high in the ionosphere Field due to a proton at the location of the electron in the H atom. The radius of the electron orbit is 0.510-10 m. E = kq /r2 = 1010 1.610-19/ (0.5 10-10 )2 = 41011 N /C + – r Hydrogen atom . Earth – – – – – – – – – E +++++ 1N / C = Volt/meter Example of field lines as long as a uni as long as m distribution of positive charge on one side of a very large nonconducting sheet. How would the electric field change if both sides were charged How would things change if the sheet were conducting This is called a uni as long as m electric field. Methods of evaluating electric fields Direct evaluation from Coulombs Law or brute as long as ce method If we know where the charges are, we can find E from E = kqi/ri2. This is a vector equation in addition to can be complex in addition to messy to evaluate in addition to we may have to resort to a computer. The principle of superposition guarantees the result. Instead of summing the charge we can imagine a continuous distribution in addition to integrate it. This distribution may be over a volume, a surface or just a line. E = dE = kdqi/r2 r where r is a unit vector directed from charge dq to the field point. dq = dV , or dq = dA, or dq = dl

Example of finding electric field from two charges Find x in addition to y components of electric field due to both charges in addition to add them up We have q1=+10 nC at the origin, q2 = +15 nC at x=4 m. What is E at y=3 m in addition to x=0 point P P Use principle of superposition Example continued Field due to q1 E = 1010 N.m2/C2 10 X10-9 C/(3m)2 = 11 N/C in the y direction. Recall E =kq/r2 in addition to k=8.99 x 109 N.m2/C2 Ey= 11 N/C Ex= 0 Field due to q2 5 E = 1010 N.m2/C2 15 X10-9 C/(5m)2 = 6 N/C at some angle f Resolve into x in addition to y components f E Ey=E sin f = 6 3/5 =18/5 = 3.6 N/C Ex=E cos f = 6 (-4)/5 =-24/5 = -4.8 N/C f Now add all components Ey= 11 + 3.6 = 14.6 N/C Ex= -4.8 N/C Magnitude Example continued Ey= 11 + 3.6 = 14.6 N/C Ex= -4.8 N/C E Magnitude of electric field f1 f1 = atan Ey/Ex= atan (14.6/-4.8)= 72.8 deg Using unit vector notation we can also write the electric field vector as:

Example of two identical charges on the x axis. What is the filed on the y axis Ey=2E sin f = 26 3/5 = 7.2 N/C Ex=0 E = 1010 N.m2/C2 15 X10-9 C/(5m)2 = 6 N/C q Ey=0 Ex=2E cos f = 26 4/5 = – 9.6 N/C Ex Ey Example of two opposite charges on the x axis. What is the filed on the y axis 4 equal charges symmetrically spaced along a line. What is the field at point P (y in addition to x = 0) q3 q2 q1 q4 P Find electric field due to a line of uni as long as m + charge of length L with linear charge density equal to l q y +x dq r dEy= dE cos q -x dE = k dq /r2 dEy dq = ldx dEy= k l dx cos q /r2 Ey= k l q cos q /r2 as long as a point charge y x -L/2 L/2 0 x= y tanq dx = y sec2 q dq r =y sec q r2 =y2sec2 q dx/r2 = dq/y dEx dE

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 y =10 cm . Typical field as long as the electrostatic smoke cleaner Ey +++++++++++++++++++++++++++++++++++++++++ Electric field gradient When a dipole is an electric field that varies with position, then the magnitude of the electric as long as ce will be different as long as the two charges. The dipole can be permanent like NaCl or water or induced as seen in the hanging pith ball. Induced dipoles are always attracted to the region of higher field. Explains why wood is attracted to the teflon rod in addition to how a smoke remover or microwave oven works. Show smoke remover demo. Electrostatic smoke precipitator model Negatively charged central wire has electric field that varies as 1/r (strong electric field gradient). Field induces a dipole moment on the smoke particles. The positive end gets attracted more to the wire In the meantime a corona discharge is created. This just means that induced dipole moments in the air molecules cause them to be attracted towards the wire where they receive an electron in addition to get repelled producing a cloud of ions around the wire. When the smoke particle hits the wire it receives an electron in addition to then is repelled to the side of the can where it sticks. However, it only has to enter the cloud of ions be as long as e it is repelled. It would also work if the polarity of the wire is reversed

Electric Dipoles: A pair of equal in addition to opposite charges q separated by a displacement d is called an electric dipole. It has an electric dipole moment p=qd. P r IEI ~ 2kqd/r3 when r is large compared to d p=qd = the electric dipole moment IEI ~2kp/r3 Note inverse cube law IEI = kq/r2 Note inverse square law as long as a single charge. + – d p +q -q Water (H2O) is a molecule that has a permanent dipole moment. When a dipole is an electric field, the dipole moment wants to rotate to line up with the electric field. It experiences a torque. GIven p = 6.2 x 10 – 30 C m And q = -10 e in addition to q = +10e What is d d = p / 10e = 6.2 x 10 -30 C m / 101.6 x 10 -19 C = 3.9 x 10 -12 m Very small distance but still is responsible as long as the conductivity of water. Leads to how microwave ovens heat up food H2O in a Uni as long as m Electric Field There exist a torque on the water molecule To rotate it so that p lines up with E. x Torque about the com = t F x sin q + F(d-x)sin q = Fdsin q = qEdsin q = pEsin q = p x E Potential Energy = U = -W = -pEcosq = – p. E Is a minimum when p aligns with E t = p x E

Motion of point charges in electric fields When a point charge such as an electron is placed in an electric field E, it is accelerated according to Newtons Law: a = F/m = qE/m as long as uni as long as m electric fields a = F/m = mg/m = g as long as uni as long as m gravitational fields If the field is uni as long as m, we now have a projectile motion problem- constant acceleration in one direction. So we have parabolic motion just as in hitting a baseball, etc except the magnitudes of velocities in addition to accelerations are different. Replace g by qE/m in all equations; For example, In y =1/2at2 we get y =1/2(qE/m)t2 Example: An electron is projected perpendicularly to a downward electric field of E= 2000 N/C with a horizontal velocity v=106 m/s. How much is the electron deflected after traveling 1 cm. Since velocity in x direction does not change, t=d/v =10-2/106 = 10-6 sec, so the distance the electron falls upward is y =1/2at2 = 0.5eE/mt2 = 0.51.610-192103/10 – 30(10-8)2 = 0.016m V E d e E Demo Transparent CRT with electron gun Back to computing Electric Fields Electric field due to a line of uni as long as m charge Electric field due to an arc of a circle of uni as long as m charge. Electric field due to a ring of uni as long as m charge Electric field of a uni as long as m charged disk Next we will go on to another simpler method to calculate electric fields that works as long as highly symmetric situations using Gausss Law.

Field due to arc of charge dEx= k dq cos q /r2 s=r q ds=r dq dEx= k l ds cos q /r2 What is the field at the center of a circle of charge Ans. 0 Find the electric field on the axis of a uni as long as mly charged ring with linear charge density l = Q/2pR. dq = lds r2 =z2+R2 cos q =z/r =0 at z=0 =0 at z=infinity =max at z=0.7R Warm-up set 2 1. [153709] Can there be an electric field at a point where there is no charge Can a charge experience a as long as ce due to its own field Please write a one or two sentence answer as long as each question. 2. [153707] An insulator is a material that three are correct is not penetrated by electric fields none of these cannot carry an electric charge cannot feel an electrical as long as ce 3. [153708] Which of the following is true of a perfect conductor There can be no electric charge on the surface. There cannot be an electric field inside. There cannot be any excess electric charge inside. There cannot be any electric charges inside. Two of the choices are correct

Kelvin Water Drop Generator Am. J. Phys. 68,1084(2000)

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