Electric Fields And Electric Fields Gauss’s Law

ElectricElectricFieldsFieldsandGauss’s LawJanuary 15, 2014Physics for Scientists & Engineers 2, Chapter 221

Section 2 missing clicker IDs! If your clicker ID is on this list, see me after class or emailme your clicker ID:! #01A0BC1D #07625336 #086EC2A4 #0AA775D8#1F6B8FFB #26120236 #333DE4EA #389BEE4D#81752CD8 #82916D7E #82A87B51 #82BE6A56#842C1FB7 #844E27ED #85319E2A #88C53D70#890B2FAD #8937DC62 #8ABF9AAF #8B9E9580#92F28BEB #92F4BADC #93002CBF #930D8C12#932F58E4 #9345D006 #95308326 #95349534January 15, 2014Physics for Scientists & Engineers 2, Chapter 212

Clicker Quiz! Three charges are arranged on a straight line as shown in thefigure! What is the direction of the electrostatic force on theright charge? (Note that the left charge is double what it wasin the previous clicker quiz)January 15, 2014Physics for Scientists & Engineers 2, Chapter 214

The Electric Field! The electric field is defined at any point in space as the netelectric force on a charge, divided by that charge F (r )E (r ) q! Electric field linesstart at positivecharges and end atnegative charges! Electric field line density: Test charge q High density of field lines – large field Low density of field lines – small fieldJanuary 15, 2014Physics for Scientists & Engineers 2, Chapter 225

Superposition of electric fields! The total electric field at any point is the vector sum of allindividual electric fields Enet Eii! Note that this means adding the x-, y-, and z- componentsseparatelyEnet ,X Ei ,XiEnet ,Y Ei ,YiEnet ,Z Ei ,ZiJanuary 15, 2014Physics for Scientists & Engineers 2, Chapter 216

Force Due to an Electric Field! The force exerted by an electric field on a point charge is F qE! The force vector is alwaystangent to the electric fieldlines and points in thedirection of the electricfield for a positive charge! The force on a negativecharge would be in theopposite directionJanuary 15, 2014Physics for Scientists & Engineers 2, Chapter 227

Electric Dipole in Electric Field! Place a dipole in a uniform electricfield! The dipole is composed of twocharges, q and –q, separated by adistance d! The net force on the dipole is zero! The electric force on the twocharges produces a torque given by τ r F, τ rF sin θ qEd sin θ , r d! Which can be written as τ pE sinθ or τ p E! The direction is given by theright-hand ruleJanuary 15, 2014Physics for Scientists & Engineers 2, Chapter 2210

General Charge Distributions! Line of charge: charge dq distributed over length dx:charge densitydqλ dx! Sheet of charge: charge dq distributed over areadx dy dA: charge densitydqσ dA! Volume of charge: charge dq distributed over volumedx dy dz dV: charge densitydqρ dVJanuary 15, 2014Physics for Scientists & Engineers 2, Chapter 2213

Electric field from charge distribution! The magnitude of the electric field at a point P due to aninfinitesimal charge dq is dq dE k 2r dq dE k 2 r̂r! Where is a vector pointing to P with magnitude 1.! The total field is given by the integral Enet dE dqP k 2 r̂P rPall points P! The integral is solved by transforming from an integral overdq to an integral over the position P (length, area, volume)January 15, 2014Physics for Scientists & Engineers 2, Chapter 2214

Finite Line of Charge! To find the electric field along a line bisecting a finite lengthof wire with linear charge density λ, we integrate thecontributions to the electric field from all the charge! We assume that the wire lies along the x-axisJanuary 15, 2014Physics for Scientists & Engineers 2, Chapter 2215

Finite Line of Charge! The symmetry of the situation allows us to conclude thatthere cannot be any electric force parallel to the wire! We can calculate the electric field due to all the charge forx 0 and multiply the result by 2! Consider a differential charge dq on the x-axis! The magnitude of the the electric field dE at a point (0,y) dueto this charge isdE kdqr2r x2 y2! The component of the electric fieldperpendicular to the wire isdqdq y dqydE y k 2 cosθ k 2 k 3rr r rJanuary 15, 2014Physics for Scientists & Engineers 2, Chapter 2216

Finite Line of Charge! Taking dq λdx, the differential electric field isdE y kλdxyr3! The electric field at a distance y from the wire is thenaaE y 2 dE y 2 k00aλdxy 2kλ y 3r0(adx2x y2)3 2kλ y 0(xdx2 y)2 3/2! The integral isa (x0dx2 yJanuary 15, 2014)2 3/2 1 2 ya 1 222x y 0 yxay 2 a2Physics for Scientists & Engineers 2, Chapter 2217

Finite Line of Charge! The electric field is then 1E y 2kλ y 2 y 2kλ 22yy a aay 2 a2! When the wire becomes infinitely longa a2y a2 1! The electric field a distance y from an infinitely long wire is2kλEy yJanuary 15, 2014Physics for Scientists & Engineers 2, Chapter 2218

Ring of ChargePROBLEM! Consider a charged ring withradius R 0.250 m! The ring has uniform chargedensity and total charge ofQ 5.00 μC! What is the electric field atd 0.500 along the axis?SOLUTIONTHINK! The charge is evenly distributed around the ring! We can calculate the electric field by integrating thedifferential field to do the differential charge! By symmetry the field will be parallel to the axis of the ringJanuary 15, 2014Physics for Scientists & Engineers 2, Chapter 2219

Ring of ChargeSKETCH! We define an x-ycoordinate systemRESEARCH! The differential electric field, dE, at x d is due a differentialcharge dq located at y R! The distance from (d,0) to (0,R) isr R2 d 2! The magnitude of the differential electric field isdqdE k 2rJanuary 15, 2014Physics for Scientists & Engineers 2, Chapter 2220

Ring of Charge! The magnitude of the x-component isddEx dE cosθ dErSIMPLIFY! The total electric field is obtained by integrating dEx over allthe charge in the ringd dqEx dEx k 2r rringring! We need to integrate around the circumference of the ringof charge using the differential arc length dsQdq ds2π RJanuary 15, 2014Physics for Scientists & Engineers 2, Chapter 2221

Ring of Charge! Now we write the integralEx 2π R 0 Q d kQd k ds 3 2π R r 2π Rr3 2π R ds 0kQdkQd 22 3/2r3(R d )CALCULATE! Putting in our numerical values8.99 10( 9ExN m2 /C2 )(5.00 10 6 C )( 0.500 m ) ( 0.250 m ) ( 0.500 m ) 22 128,654 N/CROUND! We round our results to three significant figuresEx 1.29 105 N/CJanuary 15, 2014Physics for Scientists & Engineers 2, Chapter 2222

Ring of ChargeDOUBLE-CHECK! Our units of N/C are correct! Let’s calculate the field for d REx (R2kQd d)2 3/2 kQd(d )2 3/2kQd kQ 3 2dd! Which is the result for a point charge, so our result seemsreasonable! Now calculate the field for d 0Ex (R2kQd d)2 3/2 0! Which is what we would expect at the center of a ring ofchargeJanuary 15, 2014Physics for Scientists & Engineers 2, Chapter 2223

January 15, 2014 Physics for Scientists & Engineers 2, Chapter 22 5 The Electric Field ! "e electric !eld is de#ned at any point in space as the net electric force on a charge, divided by that charge ! Electric #eld lines