mughawish (am72963) – HW 1 – coker – (56000) 1This print-out should have 12 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.This assignment covers the material fromChapter 20 in the text: Coulomb force, elec-tric field, direct integration over charge distri-butions, the dipole, properties of the E field,and electrical properties of materials.001 10.0 pointsA particle of mass 17 g and charge 18 µC i sreleased from rest when it is 21 cm from asecond particle of charge −17 µC.Determine the magnitude of the initial ac-celeration of the 17 g particle.Correct answer: 3668.37 m/s2.Explanation:Let : m = 17 g ,q = 18 µC = 1.8 ×10−5C ,d = 21 cm = 0.21 m ,Q = −17 µC = −1.7 ×10−5C , andke= 8.9875 × 109N ·m2/C2.The force exerted on the particle isF = ke|q||Q|d2= m aa = ke|q||Q|m d2= (8.9875 × 109N · m2/C2)×1.8 × 10−5C−1.7 × 10−5C(0.017 kg) (0.21 m2)= 3668.37 m/s2.002 10.0 points[This question will help you review both basicvector concepts and basic calculus concepts.]Charge Q is on the y axis a distance a fromthe origin and charge q is o n the x axis adistance d from the ori gin.What is the value of d for which the xcomponent of the force on q is the greatest?1. d =qQa√22. d =√2 a3. d =qQa24. d = a5. d =qQa6. d =qQ√2 a7. d = 08. d =a29. d =a√2correctExplanation:We have the force on charge q o n the x axisdue to charge Q on the y a xis:~F =14 π ǫ0q Qr2ˆr ,where r =pa2+ d2, so the x component ofthe force on q isFx=14 π ǫ0q Qr2cos θ=14 π ǫ0q Qa2+ d2d√a2+ d2=14 π ǫ0q Q d(a2+ d2)3/2.For maximum x component of the force,∂ Fx∂d= 0 is requir ed, so∂Fx∂d=q Q4 π ǫ0a2− 2 d2(a2+ d2)5/2= 0a2− 2 d2= 0d =a√2.003 10.0 points[A tiny charged sphere behaves like a pointmughawish (am72963) – HW 1 – coker – (56000) 2charge when you are well away from thesphere.]Two small spheres carry electric charges ofequal mag ni tudes. There are equally spacedpoints (a , b , and c) which lie along the sameline.abc− −What is the direction of the net electric fieldat each point due to these charges?1.abc− −correct2.abc− −3.abc− −4.abc− −5.abc− −6.abc− −7.abc− −8.abc− −9.abc− −10.abc− −Explanation:Since the field originates from positivecharges and terminates on the negativecharges,abc− −004 (part 1 of 2) 10.0 pointsTwo point charges, 2.6 pC and −2. 6 pC,are separated by 7 µm.What is the dipole moment of this pair ofcharges?Correct a nswer: 1. 82 × 10−17C · m.Explanation:Let : q1= 2.6 pC = 2.6 × 1 0−12C ,q2= −2.6 pC = −2.6 × 1 0−12C , andL = 7 µm = 7 × 10−6m .The dipole moment is ~p = q~L , sop = q L = (2.6 × 10−12C)(7 × 10−6m)=1.82 × 10−17C · m .005 (part 2 of 2) 10.0 pointsSketch the pair of charges, including the di-rection o f the dipole moment.1.−q +q~p2.−q +q~p3.−q +q~p4.−q +q~pcorrectmughawish (am72963) – HW 1 – coker – (56000) 3Explanation:006 10.0 pointsA dipole (electrically neutral) is placed in anexternal field.− +(a)−+(b)− +(c)−+(d)For which sit uation(s) shown above is thenet force on t he dipole equal to zero?1. (b) and (d)2. (b), (c), and (d)3. (c) and (d) correct4. (a), (b), and (c)5. (a) only6. No ne of these7. (c) only8. (a) and (d)9. Anot her combination10. (a ) and (c)Explanation:The force o n a charge in the electric field isgiven by~F = q~E a nd the torque is defined as~T = ~r ×~F .∆~E =k ∆qr2ˆr and~E =X∆~Ei.Symmetry of the configuratio n will causesome component of the electric field to bezero.Gauss’ law statesΦS=I~E · d~A =Qǫ0.The electric dipole consists of two equal andopposite charges separated by a distance. Ineither situation (c) or ( d), the electric fieldis uniform and para llel everywhere. Thus,the electric force on one charge i s equal butopposite to that on another so that the netforce on the whole dipole i s zero. By contrast,electric fields are nonuniform for situationsboth (a ) and (b).007 10.0 points[Remember that dE = kdqbr/r2. You have todo the integral. ]A uniformly charged insulating rod oflength 18.5 cm is bent into the shape of asemicircle as in t he figure.18.5−6.45 µCOIf the rod has a total charge of −6.45 µC,find the horizontal component of the electricfield at O, the center of the semicircl e. Defineright as positive. The value of the Coulombconstant is 8.98 7 55 × 109N · m2/C2.Correct a nswer: −1.06 423 × 107N/C.Explanation:Let : q = −6.45 µC = −6.45 × 10−6C ,L = 18.5 cm = 0.185 m , andk = 8.98755 × 109N · m2/C2.mughawish (am72963) – HW 1 – coker – (56000) 4byxθCall the length of the rod L and its chargeq. Due to symmetryEy=ZdEy= 0 andEx=ZdE sin θ = keZdq sin θr2,where dq = λ dx = λ r dθ, so thatEx= −keλrZ3π/2π/2cos θ dθ= −keλr(sin θ)3π/2π/2= 2keλr,where λ =qLand r =Lπ, andEx=2 keq πL2=2 (8.98755 × 109N · m2/C2)(0.185 m)2× (−6.45 × 10−6C) π=−1.06423 × 107N/C .Since the rod has negative charge, a posi tivetest charge at O would feel an attractive forcefrom the semicircle (pointing to the left) andthe field points to the left (toward the chargedistribution).008 10.0 points[Remember that dE = kdqbr/r2. You have todo the integral.]A line of charge starts at x = x0, where x0is positive, and extends along the x-axis topositive infinity.If the linear charge density is given by λ =λ0x0/x, where λ0is a positive constant, findthe electric field at the origin. (ˆı denotes theunit vector in the posi tive x direction.)1.k λ02 x20(ˆı)2.k λ02 x0(−ˆı) correct3.k λ02 x0(ˆı)4.k λ202 x0(ˆı)5.k λ0x0(−ˆı)6.k λ0x0(ˆı)Explanation:This is a continuous distribution o f charge(as opposed to point charges). We must di-vide the distribution into small elements andintegrate. Using Coulomb’s law, the elec-tric field created by each small element withcharge dq isdE =k dqx2wheredq = λ dx =λ0x0xdx , so|~E| =Zk dqx2=Z∞x0k λ0x0dxx3= −k λ0x021x2∞x0=k λ02 x0.Since the distribution is to the right of thepoint of interest, the electric field is directedalong the −x axis if λ0is po si tive. That is, apositive charge a t the o r igin would experiencea force in the direction of −ˆı from this chargedistribution. In fact, the direction of an elec-tric field at a point P in space is defined asthe direction in which the electric force actingon a positive particle at that point P wouldpoint:~E =k λ02 x0(−ˆı).mughawish