Physics 00: Sample Problems for Eam 1 1. Two particles are held fied on the -ais. The first particle has a charge of Q 1 = 6.88 10 5 C and is located at 1 = 4.56 m on the -ais. The second particle has a charge of Q =.33 10 4 C and is located at = 8.05 m on the -ais. A third particle of charge q < 0 and mass m = 0.765 g is located at +.89 m on the -ais and is free to move. (a) If charge q eperiences a net electric force of +5.56 10 4 N at this location from the fied charges, what is the charge on this particle? (b) At what acceleration will this charge q move at this location? (Show all work!) Q 1 > 0 q < 0 Q > 0 1 0 q F 1q F q Part (a): Since Q 1 > 0 and q < 0, the force (F 1q ) between Q 1 > 0 and q < 0 points to the left (as shown in the figure above) due to the fact that these charges will attract each other. Since Q > 0 and q < 0, the force (F q ) between Q > 0 and q < 0 points to the right, once again due to the fact that these charges will attract each other. As stated in the problem, the sum of these two forces is +5.56 10 4 N (F tot ). Using this fact in conjunction with Coulomb s Law, we can do a little algebra to solve for q: F tot q = F q F 1q = k e Q q = F tot k e r q [ Q r q [ Q 1 q Q k e = k e q Q 1 r 1q ] 1. r q Q ] 1, Q 1 Q rq r 1q r q = q 1 =.89 m ( 4.56 m) = 7.45 m = ( q ) = 8.05 m.89 m = 5.16 m = 6.88 10 5 C (7.45 m) = 6.88 10 5 C 55.505 m = 1.4 10 6 C/m. =.33 10 4 C (5.16 m) =.33 10 4 C 6.656 m = 8.75 10 6 C/m. 5.56 10 4 N [ q = 8.75 10 6 8.99 10 9 N m /C C/m 1.4 10 6 C/m ] 1 = 8.4 10 9 C. Since q has to be negative as stated in the problem, q = 8.4 10 9 C. Part (b): Here we just use Newton s Second Law of Motion: F = m a, and solve for a the acceleration (remembering to change the mass from cgs units to SI units): a = F tot m = 5.56 10 4 N 7.65 10 4 kg = 0.77 m/s.
PHYS-00: General Physics II Sample Problems 1 Page. A proton is released from rest at = 4.56 m in a uniform electric field of +6.68 10 6 N/C ˆ. (a) Calculate the change in potential energy when the proton moves along the -ais to the = +7.58 osition. (b) What will be the velocity of the proton at this position of +7.58 m? (c) What is the acceleration of this proton as it moves through the electric field? (Show all work!) E = 6.68 10 6 N/C Part (a): The figure to the right diagrams the problem, where our particle (a proton), q = +e has mass = 1.67 10 7 kg, charge m q = +e = 1.60 10 19 C, and travels along p o 0 v the -ais from = -4.56 m (where it starts from rest) to f = 7.58 m for a total distance of d = f = 7.58 m - (-4.56 m) = 1.14 m. This proton travels in an electric field E of strength 6.68 10 6 N/C which runs parallel to the ais. The change in potential energy can now be determined with: PE = q E d = (1.60 10 19 C)(6.68 10 6 N/C)(1.14 m) = 1.30 10 11 J, where the negative sign results from the fact that q > 0 (i.e., a positive value). f Part (b): To determine the final velocity, the simplest way is to just use the conservation of energy. Here, we do not need to include the gravitational potential energy since the mass just moves in the horizontal direction, hence PE g i = PE g f. Since we are starting from rest, the initial kinetic energy is equal to 0. KE i + PE i = KE f + PE f PE i PE f = KE f KE i PE = 1 m v f 0 v f v f = = ( PE) ( 1.30 10 11 J) 1.67 10 7 kg v f = 1.5 10 8 m/s.
PHYS-00: General Physics II Sample Problems 1 Page 3 Part (c): There are a few different ways we could solve this problem, I will show two of them here. The first. and easiest, just involves making use of Newton s Second Law of Motion (F = m a), the second involves making use of one of the equations of one dimensional motion. Newton s Second Law solution: a = F e = q E = e E = (1.60 10 19 C)(6.68 10 6 N/C) 1.67 10 7 kg = 6.40 10 14 m/s One-dimensional equation of motion solution: v = v + a ( ) = v + a d a d = v v = v 0 = v a = v d = (1.5 108 m/s) (1.14 m) = 6.44 10 14 m/s Both solutions are close to each other, but not eact. This results from round-off errors in the onedimensional equation of motion solution. The Newton s Second Law solution should be considered more eact since it uses numeric values supplied to the user instead of values that were calculated for the input values.
PHYS-00: General Physics II Sample Problems 1 Page 4 3. In the figure below, let Q 1 = +6.84 µc, Q = -16. µc, and Q 3 = +9.0 µc be rigidly fied and separated by r 1 = r 13 = r 3 = 1.34 m. Calculate the electric potential at point P which is half-way between Q 1 and Q. (Show all work!) Q 3 y First we need to calculate the distances that point P is from each charge. We will define r 1 as the distance from Q 1 to P, r as the distance from Q to P, and r 3 as the distance from Q 3 to P note that r 3 is perpendicular to r 1 : r 13 r 3 r 1 = 1 r 1 = 1 (1.34 m) = 0.670 m r = 1 r 1 = 1 (1.34 m) = 0.670 m r 3 = r3 r = 1.16 m Q P 1 Q r 1 Now we use the superposition principle (Eq. II-8) to calculate the total potential at point P: V = k e 3 i=1 ( ) qi r i ( Q1 = k e + Q + Q ) 3 r 1 r r 3 = ( [ 8.99 10 9 N m /C ) 6.84 10 6 C 0.670 m + 16. 10 6 C + 9.0 10 6 C 0.670 m 1.16 m = ( 8.99 10 9 N m /C ) [ 6.04 10 6 C/m ] = 5.43 10 4 V. ]
PHYS-00: General Physics II Sample Problems 1 Page 5 4. A fully charged capacitor stores 6.89 10 6 ev of energy while connected to a 1.50 V battery. (a) If this capacitor has circular plates of radius 1.34 cm, what is the area on one of the plates? (b) What is the capacitance of this capacitor? (c) What must be the separation between these two plates (in mm)? (d) How much charge is on these plates when fully charged? (Show all work including units!) Part (a): Since there are no other components in this circuit, the potential across the plates of the capacitor is the same as the potential delivered by the battery: V = 1.50 V. We now need to convert the energy of this capacitor from evs to SI units: U = 6.89 10 6 ev 1.60 10 19 J/eV = 1.10 10 1 J. To calculate the area, we first must convert the radius from cgs to SI units, r = 1.34 cm = 1.34 10 m. For a circular area, we get A = π r = π (1.34 10 m) = 5.64 10 4 m. Part (b): We determine the capacitance using the capacitor energy equation (Eq. II-8): U = 1 C ( V ) U C = ( V ) = (1.10 10 1 J) = 9.81 10 13 F = 0.981 pf. (1.50 V) Part (c): For the separation use Eq. (II-14): A C = ɛ d A d = ɛ C = ( 8.85 10 1 C /N m ) 5.64 10 4 m 9.81 10 13 F = 5.09 10 3 m = 5.09 mm. Part (d): We can now use either Eq. (II-1) or Eq. (II-8) to calculate the full charge, here we will use Eq. (II-1) first, then Eq. (II-8): C = Q V Q = C V = (9.81 10 13 F)(1.50 V) = 1.47 10 1 C. U = 1 Q V Q = U V = (1.10 10 1 J) 1.50 V = 1.47 10 1 C.