Physics 9 Fll 0 Homework - s Fridy September, 0 Mke sure your nme is on your homework, nd plese box your finl nswer. Becuse we will be giving prtil credit, be sure to ttempt ll the problems, even if you don t finish them. The homework is due t the beginning of clss on Fridy, September 9th. Becuse the solutions will be posted immeditely fter clss, no lte homeworks cn be ccepted! You re welcome to sk questions during the discussion session or during office hours.. Consider n infinite number of identicl chrges, ech of chrge q, plced long the x xis t distnces,, 3, 4,, from the origin. Wht is the electric field t the origin due to this distribution? Hint: use the identity + + 3 + = π 6. The net electric field is just the sum of the fields from ech chrge. So, since ech field is given by the Coulomb lw, then the net chrge is E net = + ) + 3) + 4) +, which we cn write s E net = + + 3 + +. 4 According to the hint we cn write the term in brckets s π /6, nd so the totl field is E net = π 6.
. A chrge of 3.00 µc is uniformly distributed on ring of rdius 0 cm. Find the electric field strength on the xis t distnces of ).0 cm, b) 0.0 cm, nd c) 5.0 m from the center of the ring. d) Find the field strength 5.0 m using the pproximtion tht the the ring is point chrge t the origin, nd compre your results for prts c) nd d). Is your pproximtion good one? Explin your nswer. The electric field of chrged ring of chrge Q nd rdius R is E z) = zq z + R ). 3/ We just need to plug in the distnces long z. ) Tking z = 0.0 m, then E z) = zq z + R ) = 9 09 0.0 3 0 6 =.66 0 5 N/C. 3/ 0.0 + 0. ) 3/ b) Next, with z = 0. m, we hve E z) = zq z + R ) = 9 09 0. 3 0 6 = 9.55 0 5 N/C. 3/ 0. + 0. ) 3/ c) Finlly, with z = 5 m, we find E z) = zq z + R ) = 9 09 5 3 0 6 = 079 N/C. 3/ 5 + 0. ) 3/ d) Treting the ring like point chrge we just hve Coulomb s lw, E z) = q z = 9 09 3 0 6 5 = 080 N/C, which is extremely close to the ctul nswer found in prt c). Thus, this pproximtion is excellent t these lrge distnces. This mkes sense since the detils of the chrge distribution re not seen t lrge distnces, nd ll we see is bsiclly point chrge.
3. Show tht the electric field strength E on the xis of ring chrge of rdius R hs mximum vlues t z = ±R/. Wht is this mximum vlue of E? zq z +R ) 3/, The electric field long the xis of ring of chrge Q is given by E z) = where z is the distnce long the xis, nd R is the rdius of the ring. So, the field depends on z. It hs mximum when de dz ) de dz = Q d dz z z + R ) 3/ = Q = 0. Tking the derivtive gives ) z + R ) 3 z = 0 3/ z + R ) 5/ This is zero when the term inside the prenthesis vnishes. This gives z + R = 3z, or z = ± R, s climed. When z = ± R, then E ± R ) = ) ±R/ Q ±R/ ) ) 3/ = ± + R RQ 3R /) = ± 3/ 3 Q 3 R Thus, the mximum vlue of the electric field is E = ± 3 Q 3. R 3
4. A chrge +q of mss m is free to move long the x xis. It is in equilibrium t the origin, midwy between pir of identicl point chrges, +Q, locted on the x xis t x = + nd x =. The chrge t the origin is displced smll distnce x nd relesed. Show tht it cn undergo simple hrmonic motion with n ngulr frequency 4Q ω 0 = m. 3 Hint: use the binomil expnsion. When the chrge q is t the origin, it is in stble equilibrium since the net force on it is zero. The chrge Q t x = pushes q to the right, while the chrge Q t x = + pushes it bck to the left, both with equl mgnitude. So, t equilibrium, F net = Q Q = 0. Now, suppose tht we shift the chrge q to the right by smll mount x. Then the force from the left is little bit smller, while the force from the right is little bit bigger. Hence, the net force is which cn be rewritten s F net = F net = Q Q + x) Q x), + x ) x ). Now, since x, we cn expnd this nswer s F net Q x x But, the force is F = m = m d x dt, so the ccelertion is d x dt = 4Q m 3 x, = 4Q 3 x. which is the eqution for simple hrmonic oscilltor with ngulr frequency 4Q ω 0 = m. 3 4
5. The field of n electric dipole decreses s /r 3 when the distnce of given point to the dipole, r, is much lrger thn the seprtion between the chrges. The only wy to rrnge two chrges with totl chrge of zero is to form dipole. There re, however, mny wys to rrnge four chrges with totl chrge of zero in compct pttern. An rrngement with n electric field tht behves t gret distnces s /r 4 is n electric qudrupole. For four chrges ligned with lternting signs such s + +, so tht the combintion cts like dipoles of opposite orienttion) long n xis, show tht the field on the xis perpendiculr to the line of chrges decreses s /r 4, where r is much lrger thn ny seprtion distnce within the qudrupole. Hint: once gin, use the binomil expnsion. Consider the qudrupole seen in the figure to the right, where the drk spheres re +q, nd the light ones re q. The chrges re ll seprted by distnce, nd we re interested in the field t point r. The net field of the distribution is the sum of the fields of ech chrge. It s esy to see from the symmetry of the sitution tht the net field points entirely long the verticl direction, towrds the qudrupole. Hence we only need the verticl components. The verticl fields from the two negtive chrges re E = r + /) ) cos θ r = r + /) ), 3/ where we hve reclled tht cos θ = r/ r + /). Now, the verticl field from the two positive chrges is found in the sme wy θ θ r E + = r + 3/) ) cos θ r = r + 3/) ). 3/ So, the totl field is just the sum of these two fields, which we cn find using the binomil expnsion, E = r r +/) ) 3/ r +3/) ) 3/ = = r r +/r) ) 3/ +3/r) ) 3/ + r r 3 = 3 9 r 4r = 6 r 4, nd does go s /r 4, s dvertised. r 4r ) ) 3/ ) + 3 + 3 r ) 3 r ) ) 3/ 5