Physics 114 Exam 2 Fall 2014 Solutions. Name:


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1 Physcs 114 Exam Fall 014 Name: For gradng purposes (do not wrte here): Queston Problem Answer each of the followng questons. Ponts for each queston are ndcated n red. Unless otherwse ndcated, the amount beng spread among parts (a,b,c etc) are equal. Be sure to show all your work. Use the back of the pages f necessary.
2 1. Queston 1. (10 ponts) Consder two charged conductng spheres of rad, r 1 = 0.5 m and r = 1 m. The two spheres are connected by a 10 m long wre and are at equlbrum. The total charge on the spheres s Q = 5C and each one has charge q 1 and q respectvely. (a) How much work would be requred to move a 5 nc test charge along the wre from the surface of the small sphere to the surface of the large sphere? (b) How much work would be requred to move a 5 nc test charge from the center of the small sphere to the surface same small sphere? Compare [just say whch s bgger or are they the same] the (c) potental at the surfaces of the spheres, (d) the electrc feld at the surface of the spheres, (e) the total charge on the spheres, (f) surface charge densty on the spheres (state whch s greater sphere 1 wth radus r 1 or sphere wth radus r )? (a) The potental nsde a conductor s a constant. Thus, U = 0 and t takes no work to move the charge nsde the sphere. The electrc feld s also zero so there s no force. The answer s zero. (b) The whole system s at the same potental.. Thus, U = 0 stll and the answer s zero. (c) The two spheres are at the same potental. Charge wll move untl equlbrum s reached and everywhere n a conductor s at the same potental. (d) The electrc feld s greater at the surface of the smaller sphere, E 1 > E. Frstly, we know E s greatest at places wth the largest curvature. Secondly, we know V = V 1 so E r = E 1 r 1 so E /E 1 = r 1 /r, so E 1 > E. (e) Snce V = V 1, k Q k Q Q r, and r r Q r, so Q 1 <Q. e 1 e (f) Q 1, Q, and Q r r, so 1 > 4 r 4 r Q r r 1 1 1
3 Queston. (10 ponts) Consder the crcut below wth = 8 V, R = 4, and C = 10 F. Assume the battery s deal. Immedately after the swtch s closed (after havng been open for a long tme), (a) What s the current through each resstor? (b) What s the current through the battery (c) What s the potental dfference across the capactor (d) What s the capactance of the capactor? After the swtch has been closed for a long tme, (e) What s the current through each resstor? (f) What s the current through the battery (g) What s the potental dfference across the capactor (h) What s the capactance of the capactor? (a) Immedately after the swtch s closed, current flows freely through the capactor. The crcut behaves as f the capactor s not there. The current through each resstor s /R = 8/4 = A (b) The current through the battery must be the sum of these currents = 4A (c) There s no chage across the capactor, V = Q/C = 0. (d) The capactance does not change. It s always 10 F. (e) Now there s no current through the resstor on the rght. The current through the one on the left s /R = 8/4 = A. (f) It s A as ths s now a sngle loop crcut as far as current s concerned. (g) The potental across the capactor must be equal to = 8 V. (h) Stll 10 F!
4 Queston 3. (10 ponts) You connect an deal battery to a sngle resstor, producng a potental dfference V across t and causng a current to flow through the resstor. Next, the resstor s removed from the crcut and cut n half crosswse (so ts length s halved). One of the halves s placed back nto the crcut, wth the battery connected to t. Compared to before the resstor was cut n half (a) What happens to ts resstance? (b) What happens to the potental dfference across the resstor? (c) What happens to the current across the resstor? (d) What happens to the power output of the battery? L (a) We have R A, and the length has decreased by a factor of, so the resstance s half of ts orgnal value. (b) The potental across the resstor s the same t comes from an deal battery. (c) Snce V = IR, and R s halved, I s doubled. (d) The power s can be expressed as IV so t doubles. It s also V /R, so t doubles.
5 Problem 1. (15 ponts) A charge of + C and mass 0 Kg moves from pont A to pont B n an electrc feld a dstance d = 5 m as shown below. The magntude of the electrc feld s 100 V/m. The charge s released from rest (ntal velocty s zero). (a) (3 ponts) What s the change n potental n gong from pont A to B? (b) (4 ponts) What s the fnal speed when t gets to pont B? (c) ( ponts) Whch pont s at a hgher potental (A or B)? (d) (3 ponts) What s the work done by the electrc feld for ths dsplacement from A to B? (e) (3 ponts) After reachng pont B, the charge s moved horzontally, perpendcular to the electrc feld, to pont C through a dsplacement of 4 m. What s the work done by the electrc feld for ths dsplacement from B to C? (a) V = Ed = (100V/m)(5m) = 500 V (b) Potental energy s converted to knetc energy. U = K. qv = ½mv. The change n potental s negatve and we have v = qv = **500 = 10 m/s m 0 (c) Pont A s at a hgher potental (E always ponts from hgh to low potental) (d) The work done s U = (qv) = 1000 J. (e) Here E s perpendcular to the dsplacement, so the work s zero.
6 Problem. (15 ponts) For the system of four capactors shown below, (a) fnd the equvalent capactance of the system, (b) the charge on each capactor, (c) the potental dfference across each capactor.
7
8 Problem 3. (15 ponts) Consder the crcut shown below. (a) Whch resstors are n seres and whch are n parallel (f any) (b) Use Krchoff s frst law to obtan one relaton between the three currents (c) Use Krchoff s second law to obtan two addtonal equatons that can be used to solve for the three currents. (d) Gven that I = amps, fnd the other two currents. (a) The two resstors are n seres. No other resstors are n seres on n parallel. (b) I1 = I + I3 (c) 13 I I= I1 5 I= 0 5 I 6 4 I3= 0 (d) Takng the thrd equaton and solvng for I we get I3 = (5* )/4 = Takng the frst equaton we have I1 = =.809
9 Possbly Useful Informaton 1 q1 q 885. X 10 F 1 ( C / N m ) r e = 1.6 X C E F q 0 q E, E = / 0 0 E. da q enc r 4 0 x = x  x1, t = t  t1 v = x / t s = (total dstance) / t v = dx/dt a = v / t a = dv/dt = d x/dt v = vo + at g = 9.8 m/s xxo = vot + (½)at r = x + yj + zk v = vo + a(xxo) r = r  r1 xxo = ½( vo+ v)t r = (x  x1) + (y  y1) j + (z  z1) k xxo = vt 1/at v = r / t, v = dr / dt a = dv / dt a = v / t U = Uf  U = W U=W V = Vf  V = W/q0 = U/q0 V = W /q0 V V f E f. ds V E. ds f V 1 40 q r Uf + Kf = U +K V V n V K = ½ mv V Es V E s x E V y E V x ; y ; z z V E 1 q1q U W s 4 0 r1 Q = CV A C 0 d l ab C 0 C 4 0 ln( b / a) b a C 4 0R Ceq C j (parallel) dq r n q r
10 1 1 C eq C (seres) Q U 1 j CV C u 1 0E C = C0 I= dq/dt 1 L R V = IR A P = IV P = I R=V /R I ( R r) Pemf = I R R (seres) 1 1 R R (parallel) eq j I = (R)e t/rc I = (QRC)e t/rc, I0 = (QRC) E = o eq q(t)= Q(1e t/rc ) j q(t) = Qe t/rc = Q/L, = Q/A, = Q/V
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