Download Solutions to Exam 1 of PHY2049 Fall 2006 in Electrical Physics and more Exams Calculus in PDF only on Docsity!
Prof. Yasu Takano Prof. Paul Avery Sep. 25, 2006
Exam 1 Solutions
- ( New problem ) Four charges each of size Q are arranged at the corners of a square of side L.
The magnitude of the net electric force (in N) on any of the charges can be written as CkQ^2 / L^2.
What is the approximate value of C?
(1) 1. (2) 1. (3) 1. (4) 0. (5) 0
Pick the charge to be in the upper right hand corner (any charge can be used since the situation is symmetric). The net force from the two adjacent charges is easily seen to be along the 45 de-
gree direction with a magnitude 2 kQ^2 / L^2 (because the charges give equal forces of kQ^2 / L^2
along the +x and +y direction). The charge across the diagonal is at a distance of 2 L , so it
provides a force of kQ^2 / 2 L^2 , along the 45 degree direction. Thus the total force is along the 45
degree direction, with a magnitude of (^) ( 2 +1/ 2 (^) ) kQ^2^ / L^2_. Thus C_ = 2 + 1/ 2 1..
You can also work out the problem directly using components along x and y. The adjacent
charges give a force of kQ^2^ / L^2 ( 1,1). The opposite charge yields a force of
( ) kQ^2^ / 2 L^2 1/ 2,1/ 2_. Thus the total force is_ ( ) kQ^2 / L^2 1 + 1/ 2 2,1 + 1/ 2 2_. When you_
square and add the components to get the total force you get 1.91 kQ^2^ / L^2_._
- ( WebAssign 22.41 ) A uniform electric field of 4.5 × 104 V/m is applied to a proton, which is initially stationary. What speed (in m/s) will the proton attain if the field accelerates the proton through a distance of 10 cm?
(1) 9.3 × 105
(2) 3.0 × 108
(3) 4.5 × 103
(4) 1.2 × 107
(5) 7.4 ×^106
The force can be written as F = eE. Therefore, the work Fd done by the electric field is eEd, where d is the distance through which the proton is accelerated. This work results in the kinetic
energy of the proton increasing from zero to^12 m vp^2. From eEd = 12 m vp^2 , you find v = 9.3 ×
105 m/s.
- ( Lecture problem with modification ) A ring of radius 2 cm has a total charge +2 × 10 −^9 C spread uniformly around the ring. A point charge − 2 × 10 −^9 C is placed at the center of the ring. What is the magnitude of the electric field in V/m at a point on the central axis 4 cm from the center?
(1) 3, (2) 11, (3) 8, (4) 2, (5) 0
The E field from a circular ring of charge is ( )
2 2 3/ 2 E ring (^) = kQz / z + R (as we derived in class,
and you can derive here from integration around the ring), pointing upwards.This gives
Ez (^) ring = 8050 V/m. The E field from the negative charge is 2 Ez (^) point = − kQ / z = − 11250 V/m.
The net E field from the ring charge and the point charge is therefore E (^) z net = − 3200 V/m.
- ( WebAssign 22.16 ) Two small charged beads are on a circular track of radius R = 2 m, as shown in the figure. Bead 1 of charge 5 μC is fixed in place; bead 2 of charge 10 μC can be
moved along the track. At what value of angle θ (in degrees) should bead 2 be positioned such
that the electric field at the center of the track is 3.2 × 104 V/m?
(1) 140 (2) 110 (3) 82 (4) 63 (5) 45
Let Q = 5 μ C. The E field (x and y components) from charge 1 is E 1 = kQ / R^2 ( 1, 0). The E field
from the second charge (charge 2Q) is E 2 = 2 kQ / R^2 ( − cos θ , −sinθ). The total field is obtained
by adding components and squaring, yielding ( ) ( )
2 2 2
E tot = kQ / R 1 − 2 cos θ + −2sin θ or
2
E tot = kQ / R 5 − 4 cos θ. Since Etot is 32000, we can solve for cos θ to get θ = 140 degrees.
R θ x
y
The force on an electron is F = eE. The potential difference between two plates is Ed. Thus the potential difference can be written V = Fd/e = 2250 V.
- ( Homework 24.48 ) Particles 1, 2 and 3 lie on a line, at positions x 1 = 0, x 2 = 4cm, and x 3 = 10cm. The charges of particles 1 and 2 are each 5 μC. The electrical potential energy of the three-particle system is 0 (where potential energy at infinity is defined to be 0). What is the charge of particle 3 in μC?
Let q be the charge of particles 1 and 2 and let q 3 be the charge of particle 3. The total potential
energy of the three particle system is U 123 (^) = kq^2 / d 12 (^) + kqq 3 (^) / d 13 (^) + kqq 3 (^) / d 23 = 0 (using obvious
notation for d 12 , d 13 and d 23. This equation yields q 3 = −4.7 μC.
- ( WebAssign 25.12 plus lecture discussion ) In the figure shown, the battery has a potential dif- ference of 24 V. If C 1 = C 4 = C 5 = 5 μF and C 2 = C 3 = 10 μF, what is the charge on C 5 in μC?
(1) 80 (2) 160 (3) 120 (4) 240 (5) 40
The total capacitance in the right hand branch is C 2345 = 3.33 μ F. The charge on this equivalent
capacitance is the same as the charge on C 5 , because C 5 is in series with the rest of the branch
C 234 (as discussed in lecture and the textbook). Thus q 5 = 3.33 × 24 = 80 μ C.
C 2 C 3
C 4
C 5
C 1
- ( Sample problem 25-6 ) A parallel-plate capacitor whose capacitance is 120 pF is charged by a battery to a potential difference of 6 V between its plates. The battery is now disconnected, and a slab of a dielectric material, whose dielectric constant is 80, is slipped between the plates. What is the potential energy of the capacitor-slab device after the slab is inserted, in joules.
(1) 2.7 × 10 −^11
(2) 5.8 × 10 −^7
(3) 1.5 × 10 −^3
(5) 7.4 × 103
When the slab is inserted, the electric field, and thus the potential, is reduced by a factor of 80.
The capacitance is increased by a factor of 80. Thus the total energy^12 CV^2 , is reduced by a fac-
tor of 80. A calculation yields U = 12 CV^2 / 80 = 12 × ( 120 × 10 −^12 ) 62 / 80 = 2.7 × 10 −^11 J.
