Version 001 Electrostatics II tubman (12125) 1. second particle of charge 13 µc. Determine the magnitude of the initial acceleration

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1 Version 001 Electrostatics II tubman (12125) 1 This print-out should have 15 questions. Multiple-choice questions ma continue on the net column or page find all choices before answering. Negativel Charged Rod points If a negativel charged rod is held near an uncharged metal ball, the metal ball 1. is unaffected. 2. Unable to determine. 3. becomes polar. correct 4. becomes negativel charged. 5. becomes positivel charged. When the charged rod moves toward the metal ball, an electric field is created around the ball. Electrons inside the metal ball can move freel, and move awa from the area where the negativel charged rod approaches, causing a positivel charged area, creating a polar metal ball. Charges points When the leaves of an electroscope are spread apart, 1. the leaves have the same charge. correct 2. the leaves are neutral. second particle of charge 13 µc. Determine the magnitude of the initial acceleration of the 12 g particle. Correct answer: m/s 2. Let : m = 12 g, q = 70 µc = C, d = 28 cm = 0.28 m, Q = 13 µc = C, and k e = N m 2 /C 2. The force eerted on the particle is q Q F = k e d 2 = ma q Q a = k e md 2 = ( N m 2 /C 2 ) C C (0.012 kg)(0.28 m 2 ) = m/s 2. Electric Field points Two charges are located in the (,) plane as shown. The fields produced b these charges are observed at a point p with coordinates (0,0). p 3. a positivel charged object must be touching the knob of the electroscope. 8.1 C 1.8 m 1.8 m 9 C 4. a negativel charged object must be touching the knob of the electroscope. Acceleration of a Particle points A particle of mass 12 g and charge 70 µc is released from rest when it is 28 cm from a 2.5 m 3 m Find the -component of the electric field at p. The value of the Coulomb constant is N m 2 /C 2. Correct answer: N/C.

2 Version 001 Electrostatics II tubman (12125) 2 Let : ( p, p ) = (0,0), ( 1, 1 ) = (3 m, 1.8 m), ( 2, 2 ) = ( 2.5 m, 1.8 m), q 1 = 9 C, q 2 = 8.1 C, and k e = N m 2 /C 2. p E = E 1 +E 2 = N/C + ( N/C ) = N/C. Field Directions b Inspection 005 (part 1 of 5) 10.0 points q q 1 Consider smmetricall placed rectangular insulators with uniforml charged distributions of equal magnitude as shown. Consider the electric field vectors: E 1 θ 2 θ 1 What is the direction of the electric field at the origin? q 2 E 2 q 1 where θ 1 = 180 tan m 3 m = , θ 2 = 180 +tan m 2.5 m = In the -direction, the contributions from the two charges are Q 1 Q 1 1 E 1 = k e r1 2 cosθ 1 = k e r1 2 r 1 = ( N m 2 /C 2) 9 C 3 m ( m) m = N/C and Q 2 Q 2 2 E 2 = k e r2 2 cosθ 2 = k e r2 2 r 2 = ( N m 2 /C 2) 8.1 C ( m) m m = N/C, so 1. Aligned with the negative -ais 2. Aligned with the positive -ais 3. Zero with undefined direction 4. Non-zero and not aligned with either the - or the -ais 5. Aligned with the positive -ais 6. Aligned with the negative -ais correct At the origin, the positive slab of charge produces an electric field pointed into quadrant III (awa from the positivel charged slab). The negativel charged slab produces an electric field of equal magnitude (as the positivel charged slab) but pointing into quadrant II (toward the negativel charged slab). The -components of the two fields add (producing E < 0), while the -components cancel, so the electric field is along the negative -ais.

3 Version 001 Electrostatics II tubman (12125) (part 2 of 5) 10.0 points at the origin? 1. Zero with undefined direction 2. Aligned with the positive -ais 3. Aligned with the negative -ais 4. Aligned with the positive -ais What is the direction of the electric field at the origin? 1. Aligned with the positive -ais 2. Aligned with the negative -ais 3. Aligned with the positive -ais 4. Zero with undefined direction correct 5. Non-zero and not aligned with either the - or the -ais 6. Aligned with the negative -ais At the origin the upper slab produces an electric field that points into quadrant III (awa from the positivel charged slab) and the lower slab produces an electric field pointing into quadrant I (awa from the positivel charged slab). B smmetr, these two electric fields are equal in magnitude and opposite in direction, so the total electric field sums to zero. 007 (part 3 of 5) 10.0 points Whatisthedirectionofthenetelectricfield 5. Non-zero and not aligned with either the - or the -ais correct 6. Aligned with the negative -ais At the origin, the upper slab produces an electric field pointing into quadrant III (awa from the positivel charged slab) while the lower slab produces an electric field also pointing into quadrant III (toward the negativel charged slab). B smmetr both slabs produce an electric field of the same magnitude and direction. The sum of these electric fields points into quadrant III which is neither aligned with the nor ais. 008 (part 4 of 5) 10.0 points What is the direction of the net field at the origin? 1. Aligned with the negative -ais 2. Aligned with the positive -ais 3. Non-zero and not aligned with either the - or -ais 4. Zero with undefined direction 5. Aligned with the positive -ais correct

4 Version 001 Electrostatics II tubman (12125) 4 6. Aligned with the negative -ais At the origin the fields from the top and bottom slabs cancel because the are equal and opposite. The field from the left slab points toward the positive -ais (awa from the positivel charged slab) and the field from the right slab also points toward the positive -ais (toward the negativel charged slab). This configuration is smmetric about the - ais, so the component of the total field must vanish, and the sum of the electric fields from all four slabs is aligned with the positive -ais. 009 (part 5 of 5) 10.0 points Whatisthedirectionofthenetelectricfield at the origin? 1. Aligned with the positive -ais 2. Aligned with the negative -ais 3. Non-zero and not aligned with either the - or the -ais 4. Aligned with the negative -ais correct 5. Aligned with the positive -ais 6. Zero with undefined direction At theoriginthefield from thebottomslab points toward the negative ais (toward the negativel charged slab). The fields from the right and left slabs have equal magnitudes at the origin and are smmetric about the ais, so their components cancel at the origin. The net fields from the right and left slabs point toward the negative ais (awa from the positivel charged slab). Thus the sum of the electric fields from all three slabs is aligned with the negative -ais. Electron in a Field (part 1 of 2) 10.0 points An electron with kg is acceleratedfromrestfor sbauniformelectric field that eerts a force of N on the electron. What is the magnitude of the electric field? The fundamental charge is C. Correct answer: V/m. Let : q e = C and F = N. The magnitude of the force is F = q e E E = F q e = N C = V/m. 011 (part 2 of 2) 10.0 points What is the speed of the electron after it has accelerated for s? Correct answer: m/s. Let : m e = kg and t = s. The force on the electron is F = m e a and its acceleration is a = v t = F m e v = F t = ( N)( s) m e kg = m/s.

5 Version 001 Electrostatics II tubman (12125) 5 AP B 1998 MC 14num points Two parallel conducting plates are connected to a constant voltage source. The magnitude of the electric field between the plates is 2762 N/C. If the voltage is quadrupled and the distance between the plates is reduced to 1 8 the original distance, what is the magnitude of the new electric field? Correct answer: N/C. Let : E = 2762 N/C, V = 4V, and d = 1 8 d. The magnitude of the electric field between two parallel conducting plates is E = V d, where V is the voltage between the plates, and d is the distance between the plates, so the new electric field has a magnitude of E = V d = 4V ( ) V = 32 d d 8 = 32E = 32(2762 N/C) = N/C. Accelerating a Deuteron 013 (part 1 of 2) 10.0 points Adeuteron(anucleusthatconsistsofoneproton and one neutron) is accelerated through a 3.09 kv potential difference. How much kineticenerg does itgain? The mass of a proton is kg, the mass of a neutron kg and the elemental charge is C. Correct answer: J. Let : V = 3.09 kv = 3090 V and q = C. The neutron has no charge. B conservation of energ since there is a gain of kinetic energ, there must be a loss of potential energ ( V < 0) and K = U = q V = ( C ) ( 3090 V) = J. 014 (part 2 of 2) 10.0 points How fast is it going if it starts from rest? Correct answer: m/s. Let : m p = kg and m n = kg. The mass of the deuteron is m = m p +m n = kg kg = kg, so K = 1 2 mv2 0 2( K) v = m 2 ( = 16 J) kg = m/s. Hewitt CP9 22 E points Would ou feel an electrical effects if ou

6 Version 001 Electrostatics II tubman (12125) 6 were inside the charged sphere of a van de Graaff generator? Wh or wh not? 1. Yes; the electric field is ver strong inside the van de Graff generator. 2. None of these 3. Yes; the electric field eists both inside and outside of the generator. 4. No; the inside of the generator has zero charge and thus no electric field. correct 5. More information is needed. 6. No; although there are charges inside the generator, the net charge is zero. You would feel no electrical effects inside an staticall charged conducting bod. The distribution of mutuall-repelling charges is such that the electric field inside the bod is zero.

Version 001 Electrostatics I tubman (12125) 1

Version 001 Electrostatics I tubman (115) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. AP EM 1993 MC 55