Physcs 114 Exam Sprng 013 Name: For gradng purposes (do not wrte here): Queston 1. 1... 3. 3. Problem Answer each of the followng questons. Ponts for each queston are ndcated n red wth the amount beng spread equally among parts (a,b,c etc). Be sure to show all your work. Use the back of the pages f necessary.
Queston 1. (10 ponts) Consder a constant electrc feld that ponts drectly downward towards the ground. (a) Compare the electrc potental at a pont P located at a heght of y = 5 m to that of a pont Q located at a heght of y = 1 m. Is pont P at a hgher, lower, or the same potental? (b) Compare the electrc potental energy of a proton located at these same ponts (P and Q). When s the electrc potental energy of the system hgher, when the proton s at P, at Q, or s the same for both? (c) Compare the electrc potental energy of an electron located at these same ponts (P and Q). When s the electrc potental energy of the system hgher, when the electron s at P, at Q, or s the same for both? (d) If the electron s moved from pont P to Q, descrbe the work done by the electrc feld. Is t postve, negatve, or zero? Solutons (a) The electrc feld ponts from hgh to low potental so the hgher pont s at a hgher potental. P s at a hgher potental. (b) The potental energy s qv, and q s postve, so the system has mre potental energy at the hgher poston. So t s hgher at P. Note, also, that you have to do work to lft the proton. (c) Now thngs are opposte. Q s negatve. The potental energy s hghest wth the electron on the ground, so t s hgher at Q. (d) The dsplacement s downward. The force by the feld on the electron s upward. Thus W = F s s negatve (theta s 180 degrees).
Queston. (10 ponts) Consder a parallel plate capactor wth an area A and plateseparaton of d wth ar between the plates. The capactor s ntally hooked up to an deal 9 V battery? (a) What happens to the capactance of the capactor f t s now hooked up to an deal 18 V battery? (b) What happens to the charge on the capactor after t s moved to the 18 V battery (compared to when t was hooked up to 9 V)? (c) How about the energy stored n the electrc feld of the capactor as t s moved from the 9V to the 18V battery? Whle the capactor s stll hooked up (and remans hooked up) to the 18V battery, you nsert a delectrc (that has a much hgher delectrc constant than ar) (d) What happens to the capactance of the capactor that now has the delectrc n t compared to before the delectrc was nserted? (e) What happens to the charge stored on the capactor after you nsert the delectrc compared to before t s nserted. (f) What happens to the energy stored n the capactor after you put the delectrc n compared to before t s put n. Solutons (a) The capactance s ndependent of the voltage across t. It remans the same. (b) Q = CV. V doubled so, Q doubles. (c) C stays the same so thnk about U = ½ CV. V doubles so U quadruples. (d) The capactance ncreases (by a factor of ). (e) Q = CV. V s constant and C ncreases so Q ncreases. (f) V stays the same so thnk about U = ½ CV. C ncreases so U ncreases as well.
Queston 3. (10 ponts) Consder the crcut below n whch there are 4 dentcal lght bulbs, labeled A, B, C, D. A B C D In each of the followng consderatons, assume only the bulb mentoned n each part s unscrewed (so n part (a)only A s unscrewed and n part (b) only C s unscrewed). (a) When A s unscrewed, what happens to the brghtness of bulbs B, C, D? (b) When C s unscrewed, what happens to the brghtness of bulb D? (c) When C s unscrewed, what happens to the brghtness of bulbs A and B? Solutons (a) When A s unscrewed, all the lghts go out snce the combnaton of B,C, D are n seres wth A. There s no complete crcut anymore. (b) Current cannot go through that path anymore (R goes way up) so D goes out. Ths s lke Chrstmas lghts and as demonstrated n class. (c) When C s unscrewed the resstance of the whole crcut ncreases (there are fewer paths for the current). Thus the current gong through A decreases and so P = I R for A decreases: A gets dmmer. The voltage drop across A also gets smaller snce I gets smaller. Ths means the voltage drop across B ncreases. Thus P = V /R ncreases for B and B gets brghter.
Problem 1. (15 ponts) A sphercal conductor has a radus of 14.0 cm and a charge of 6.0 µc. Calculate the electrc potental at the followng dstances from the center: (a) r = 10 cm, (b) r = 0 cm, (c) r = 14 cm. (d) If a partcle of mass 3 10-9 Kg and charge of nc s accelerated from rest from r = 14 cm to r = 0 cm, what wll the fnal speed of the partcle be? Solutons kq e r for r R V kq e R for r R (a) V = (9 X 10 9 )(6 X 10-6 )/0.14 = 1.67 MV (b) V = (9 X 10 9 )(6 X 10-6 )/0. = 1.17 MV (c) V = (9 X 10 9 )(6 X 10-6 )/0.14 = 1.67 MV (d) We want to employ conservaton of energy. The ntal mechancal energy = the fnal mechancal energy. K + U = K f + U f.. K = 0. K f = -( U f - U ) = -U. The change n potental energy wll be qv. K f = ½ mv V = 1.67-1.17 MV = 0.5 MV v = ( 10-9 )( 5 10 5 )/ 3 10-9 = 6.67 10 5 v = 817 m/s
Problem. (15 ponts) Consder the crcut below where are C 1 = 3F, C = 1 F, and C 3 = 5F. (a) What s the equvalent capactance f all three capactors were reduced to a sngle equvalent one? If the battery suppled an emf of 9 V, fnd (b) the charge and (c) the voltage across each of the three capactors. Soluton (a) 1/C eq = 1/ C 1 + 1/( C + C 3 ) = 1/3 + 1/6. C eq = F (b) and (c)the charge on the equvalent capactor s the same as that on C 1. Q 1 = Q eq = C eq V.= ()(9) = 18 C. The voltage across ths capactor s then V 1 = Q 1 / C 1 = 18/3 = 6 V. We can get the voltages V and V 3 by subtractng that across C 1 from the 9V so we get V = V 3 = 3 V. Now we can get the charges on these Q = C V = 3 C Q 3 = C 3 V 3 = 15 C
Problem 3. (15 ponts) The ammeter shown n the fgure below reads.00a. (a) Whch of the resstors shown are n parallel wth each other? (b) Whch of the resstors shown are n seres wth each other? Fnd (c) I 1, (d) I, (e). (a) None are n parallel. (b) None are n seres (c) Applyng Krchoff s nd law to the top loop we have 15 7I 1 (5)() = 0, so I 1 = 5/7 = 0.714 A (d) Applyng Krchoff s frst law to the left hand juncton we have I 1 + I = A, so 0.714 + I = A, and I = 1.9 A (e) Applyng Krchoff s nd law to the top loop we have ()(1.9)-(5)() = 0. Thus, = 1.6 V
Possbly Useful Informaton 1 q1 q F 0 885. X 10-1 ( C / Nm ) 4 0 r e = 1.6 X 10-19 C E F q 0 q E, E = / 00EdA. q enc 4 0r x = x - x 1, t = t - t 1 v = x/ t s = (total dstance) / t v = dx/dt a = v / t a = dv/dt = d x/dt v = v o + at g = 9.8 m/s x-x o = v o t + (½)at r = x + yj + zk v = v o + a(x-x o ) r=r -r 1 x-x o = ½( v o + v)t r = (x - x1) + (y - y1) j + (z - z1) k x-x o = vt -1/at v= r / t, v=dr /dt a = dv / dt a = v/ t U = U f - U = -W U=-W V = V f - V = -W/q 0 = U/q 0 V = -W /q 0 V V f E f. ds V E. ds f V 1 40 q r U f + K f = U +K V V n V 1 1 40 1 1 40 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 40 r1 Q = CV A C 0 d l ab C 0 C 4 0 ln( b / a) b a C 4 0R Ceq Cj (parallel) dq r n q r
1 1 C eq C Q U 1 j CV C u 1 0E C = C 0 I= dq/dt 1 L R A V = IR P = IV P = I R=V /R I ( R r) P emf = I R R (seres) 1 1 R R (parallel) eq j I = (R)e -t/rc I = (QRC)e -t/rc, I 0 = (QRC) E = o eq q(t)= Q(1-e -t/rc ) j q(t) = Qe t/rc = Q/L, = Q/A, = Q/V