Coulomb's law and electric field intensity ppt

coulomb's law electrostatics ppt and coulomb's law ppt presentation
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Published Date:23-07-2017
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Coulomb’s Law and the Electric Field Physics 2415 Lecture 2 Michael Fowler, UVa The Electroscope • Charge detector invented by an English clergyman in 1787. Two very thin strips of gold leaf hang side by side from a conducting rod. • If a + charge is brought near, electrons move up the rod, leaving the two strips positively charged, so they repel each other. Charging the Electroscope… • By conduction: touch the top conductor with a positively charged object—this will leave it positively charged (electron deficient). • By induction: while holding a positively charged object near, but not in contact, with the top, you touch the electroscope: negative charge will flow from the ground, through you, to the electroscope. Coulomb’s Law • Coulomb measured the electrical force between charged spheres with apparatus exactly like Cavendish’s measurement of G: two spheres, like a dumbbell, suspended by a thin wire. One sphere was charged, another charged sphere was brought close, the angle of twist of the wire measured the force. Coulomb’s Law • Coulomb discovered an inverse square law, like • a gravitation, except that F 21 F 12 of course like charges + + repelled each other. The Q Q 2 1 force acted along the line of centers, with _ magnitude proportional + to the magnitudes of both charges: QQ 12 Fk 2 rUnit of Charge • We can’t make further progress until we define a unit of charge. • The SI unit is the Coulomb. Its definition is not from electrostatics, but the SI unit current in a wire, one amp, is one coulomb per second passing a fixed point, and one amp is the current that exerts on an identical parallel current one meter away a magnetic force of one Newton per meter of wire. We’ll do all this later—just letting you know why we have this very large unit. Coulomb’s Law with Numbers QQ 12 Fk 2 • a r F 21 F 12 + + • Experimentally, with r in Q Q 2 1 meters and F in Newtons, it 9 is found that k = 9x10 . _ • This means that two + charges each one -3 millicoulomb (10 C), one meter apart, repel with a Note: a common 1 force of 9,000N, about one notation is k 4  0 ton weight Atomic Electrostatics • The simplest (Bohr) model of the hydrogen atom has an electron circling a proton at a -10 distance of about 0.5x10 m. • The electron charge has been determined -19 experimentally to be about -1.6x10 C. • This means the electrostatic force holding the electron in orbit QQ 12 Fk 2 r -7 is about 10 N. Atomic Dynamics • The electrostatic force holding the electron in orbit QQ 12 Fk 2 r -7 is about 10 N. -30 • The electron has mass about 10 kg, so its 23 2 acceleration is about 10 m/s . 2 6 • This is v /r, from which v is about 2x10 m/s, around 1% of the speed of light. Superposition • The total electric force on a charge Q from 3 two charges Q , Q is the vector sum of the 1 2 forces from the charges found separately. ˆ kQ Qr 2 3 23 Q F 1 23 2 Q 3 r 23 FQ on total 3 ˆ kQQr Q 1 3 13 2 F 13 2 r 13 Sounds trivial—but superposition isn’t true for nuclear forces The Electric Field Er r • The electric field at a point is defined  by stipulating that the electric force on a tiny F q test charge at is r given by F q E . • Strictly speaking, the test charge should be vanishingly small: the problem is that if the electric field arises in part from charges on conductors, introducing the test charge could cause them to move around and thus change the field you’re trying to measure. Field from Two Equal Charges • Two charges Q are placed on the y-axis, equal distances d from the origin up and down. What is the electric field at a point P on the x-axis, and where is its maximum value? y-axis Q r d E total  P x x-axis  d ˆ kQr E upper charge 2 r Q • Anywhere on the axis, the field is along the axis, and has value 2kQ 2kQx 2kQx E cos . 23 3/2 22 rr xd Field on the Axis of a Uniform Ring of Charge • Imagine the ring, radius a, total charge Q, to be made up of pairs dQ of oppositely placed charges: y-axis dQ r a E total  P x x-axis  dQ • From the previous slide, adding contributions from all pairs, kQ kQx kQx E cos . 3/2 23 22 rr xa  Visualizing the Electric Field • a • For a single point charge, we can easily draw vectors at various points indicating the strength of the field there: Visualizing the Electric Field • a • A standard approach is to draw lines of force: lines that at every point indicate the field direction there. • These lines do not immediately give the field strength, but their density can give a qualitative indication of where the field is stronger, provided they are continuous. Field from a Uniform Line of Charge • What’s the electric field at a • q point P distance R from a very z-axis long line of charge, say C/m?  dz • Take the wire along the z-axis in Rsec 3D Cartesian coordinates,we’ll R tan find the field at a point P,  distance R from the wire, in the P O R (x,y) plane. • The strategy is to find the field dE from charge z from a bit dz of the wire, dE in dR tan then do an integral over the  whole wire. Field from a Uniform Line of Charge • The strategy is to find the field dE • q z from a bit dz of the wire, then z-axis do an integral over the whole wire. dz k d R tan kdq kdz kd dE 2 2 2 2 rR sec r r R sec R zR tan • For an infinite wire, the net field  P O must be directly away from the R wire, so multiply by and cos integrate over all z : dE from charge  /2 kk  2 in dR tan  Ed cos  RR  /2Electric Field from a Line of Charge: Top View • a • It looks just like the field from a point charge: but isn’t • Remember that for a point charge Q, the magnitude of 2 field at distance R is kQ/R , for a line charge with density , the kR/2 field strength is : so the “density of lines” in these 2D plots can’t relate directly to field strength for both cases. • (Actually, if we could draw the lines in 3D, the density would relate directly to field strength.) Electric Field from a Plane of Charge: Top View Note: if you can’t follow this, it doesn’t matter Each red dot is a cross section of a wire perpendicular to the page. • a • Imagine now we have a uniformly charged plane: we make it up of many parallel dy wires, each charge density  , rR sec each perpendicular to the page, yR tan with n wires/meter.  P O • Remembering the field strength R kr/2 from a single wire is , and in dy there are ndy wires, dE from charge the field strength at P from the in dR tan charge in dy is:  2k nd R tan 2kndy dE 2knsecd RR sec secNote: the next slide is all you need to know—and it’s simple Electric Field from a Plane of Charge: Top View • We’ve shown the field strength at P from the charged lines in dy is • a 2k nd R tan 2kndy dE 2knsecd RR sec sec • This has component in the OP direction dy dE cos 2knd rR sec y The total field is given by integrating, R tan   P O E22 kn k R 2 0 n where the plane charge density dE from charge 2 Coulombs/m , and k1/ 4  . 0 in dR tan  • Notice the field strength is constant