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Phy 2049 Spring 2012 Exam 1 – v2 solutions Exam 1 Solutions 1. A graph of the x component of the electric field as a function of x in a region of space is shown in figure. The scale of the vertical axis is set by Exs = 20.0 N/C. The y and z components of the electric field are zero in this region. If the electric potential at the origin is 30 V, what is the electric potential (in V) at x = 4.0 m ? (1) -10 (2) 70 (3) 0 (4) -40
(5) 30
The change in the electric potential from the origin to x=4 is given by: x2
1
ΔV = − ∫ E dx = − Exs 4 = −40V x1
2
So the potential at x = 4 is 30 – 40 = -­‐10 V. 2. Two point-­‐like objects with electric charges q1 = +0.5 nC and q2 = -­‐0.3 nC are separated by a distance of 0.4 m. How much work (in J) is required to transport a third charge q3 = +0.2 nC from infinitely far away to a position exactly at the midpoint of line connecting the two other charges? (1) 1.8 x 10-­‐9 (2) 9 x 10-­‐9 (3) 9 (4) 3.6 x 10-­‐9 (5) 0 The electric potential at the midpoint between q1 and q2 is given by: q1
q3
V=
+
4πε 0 ( 0.2 m ) 4πε 0 ( 0.2 m )
9 × 10 9 ) ( 0.5 − 0.3)10 −9
(
V=
=9
( 0.2 m )
The work done to move q3 to this location is the change in potential energy: ΔU = q3V = 0.2 × 10 −9 C ( 9 V) = 1.8 × 10 −9 J (
)
Phy 2049 Spring 2012 Exam 1 – v2 solutions 3. In the figure shown, what is the net electric potential at point P, which is at the center of a square array of eight particles with the listed electric charges, if V=0 at infinity? (1) -­‐5q/(4π ε0 d) (2) -­‐q/(4πε0 d) (3) +q/(4πε0 d) (4) +2q/(4πε0 √ 2 d) (5) (9-­‐8/√ 2 )q/(4πε0 d) The potential energy from several point charges is the scalar sum: qi
V =∑
i 4πε 0 ri
Note that the charges on the diagonals and central vertical column cancel, leaving just the horizontal charges that don’t cancel: ( −2q − 3q ) = −5q V=
4πε 0 d
4πε 0 d
4. A proton is placed in a region where the electric potential only varies with x as shown. The scale of the vertical axes is set by VS = 500 V. What is the x component of the electric force (in N) on the proton when it is placed at x=2 m? The proton has charge q = +e = 1.6 x 10-­‐19 C. (1) 4 x 10 -­‐17 (2) -­‐4 x 10-­‐17 (3) -­‐8 x 10-­‐17 (4) 2 x 10-­‐17 (5) 0 Phy 2049 Spring 2012 Exam 1 – v2 solutions ⎛ ΔV ⎞
⎛ −V ⎞
Fx = qEx = e ⎜ −
= e ( −1) ⎜ s ⎟ = 250e = 4 × 10 −17 N ⎟
⎝ Δx ⎠
⎝ 2 ⎠
5. In the shown figure, a potential difference of V = 24 V is applied across the arrangement of capacitors with capacitances of C1 = C2 = 2μF, and C3 = 3 μF. What is the charge q2 on capacitor C2? (1) 24 μC (2) 96 μC (3) 72 μC (4) 12 μC (5) 48 μC The equivalent capacitance is −1
⎛ 1
1⎞
Ceq = ⎜ + ⎟ + C3 = 4 µF ⎝ C1 C2 ⎠
The total charge stored on the capacitors is: qtot = CeqV = ( 4 µF ) ( 24 V) = 96 µC That on C3 is similarly: qtot = CeqV = ( 3µF ) ( 24 V) = 72 µC So the difference is what is on q2, which is also what is on q1 by the way: q2 = qtot − q3 = 24 µC 6. A potential difference of V=12 V is applied across a 1 μF capacitor until it is fully charged, and then it is disconnected. If the charged capacitor is then connected in parallel with a second (initially uncharged) capacitor, and if the potential difference across the first capacitor subsequently drops to 9 V, what is the capacitance of this second capacitor? (1) 0.33 μF (2) 1.33 μF (3) 1 μF (4) 3 μF (5) 9 μF The charge on the first capacitor when first connect to 12V is: q = CV = (1µF ) (12V) = 12 µC Phy 2049 Spring 2012 Exam 1 – v2 solutions When disconnected from the potential difference and instead put in parallel with the second capacitor, this charge must remain constant. The effective capacitance has changed, however, and thus the voltage: q = 12 µC = C ′V ′ = ( C1 + C2 ) ( 9V)
4
1
⇒ C2 = µF − 1µF = µF
3
3
7. What is the equivalent capacitance between terminals a and b if each capacitor shown in the figure has capacitance C? (1) 3C/5 (2) 3C/2 (3) 5C/2 (4) C/4 (5) 4C The 2 capacitors on the vertical leg have equivalent capacitance C/2. When added to the rightmost capacitor, which is in parallel, the capacitance is 3C/2. When added to the leftmost capacitor, which is in series, the total equivalent capacitance is −1
2 ⎞
3C
⎛1
Ceq = ⎜ +
=
⎟
⎝ C 3C ⎠
5
8. A certain parallel plate capacitor with capacitance 12 μF is connected to a source of EMF with potential 3 V. A dielectric material with κ = 4 is then inserted between the plates of the capacitor. By how much does the energy stored in the capacitor change ? (1) 1.6 x 10-­‐4 J (2) 5.4 x 10-­‐5 J (3) 1.2 x 10-­‐5 J (4) 1.4 x 10-­‐6 J (5) 0 J The potential energy of capacitor is: Phy 2049 Spring 2012 Exam 1 – v2 solutions 1
U = CV 2 2
When a dielectric material is inserted into a capacitor, the new capacitance is given by: C ′ = κ C Therefore, given that the voltage does not change, the change in potential energy is given by: 1
ΔU = U ′ − U = ( C ′ − C )V 2
2
1
ΔU = C (κ − 1)V 2
2
1
ΔU = (12 µF ) ( 4 − 1) 32 = 1.6 × 10 −4 J
2
9. How much power is dissipated in a copper wire that is 15 m long and 1 mm in diameter when it is connected across a potential difference of 6 V? The resistivity of copper is ρ = 1.7 x 10-­‐8 Ω m. (1) 110 W (2) 0.32 W (3) 4 x 10 16 W (4) 12 W (5) 4 x 10 9 W 2
6 2 π ( 0.001 / 2 m )
V 2 V 2A
P=
=
=
= 111Ω R
ρL
1.7 × 10 −8 Ωm (15 m )
(
)
10. A 1800 W space heater is constructed to operate at 120 V. What is the current in the heater when the unit is operating? (1) 15 A (2) 8 A (3) 0.07 A (4) 27,000 A (5) 1 A P = iV
1800 W
i=
= 15 A
120 V
Phy 2049 Spring 2012 Exam 1 – v2 solutions 11. Three equal charges, each of +7 µC are arranged at the corners of an equilateral triangle of side length 10 cm. What is the magnitude of the total force (in N) acting on each of the charges? (a) 76 (b) 44 (c) 88 (d) 38 (e) zero d d Take for example the lower left charge in the above triangle: q2
q2
Fx = −k 2 − k 2 cos 60
d
d
2
q
Fy = −k 2 sin 60
d
2
2
q2
2
2
F = Fx + Fy = k 2 1+ cos 60 + sin 60 = 76 N
d
12. A negative charge -­‐Q and a positive charge 2Q are placed along the x-­‐axis at x=0 and x=1 m., respectively. At what point along the x-­‐axis (in meters) does the electric field equal to zero? (a) –2.4 (b) 2.4 (c) –0.4 (d) 0.4 (e) there is not such a point The position can only be two the left of the charge at x=0 because the charge at x=1 is larger. Let’s suppose that this location where the field is zero is x = −d . Then the field is given by: Q
2Q ˆ
E = +k 2 iˆ − k
i =0
d
(1+ d )2
(
2
1+ d )
(
⇒
d2
⇒1=
(
⇒d =
(
) (
=2
)
2 −1 d
1
)
2 −1
So x = −d = −2.4
)
Phy 2049 Spring 2012 Exam 1 – v2 solutions 13. A thin bar of length L is placed along the x-­‐axis as shown in Figure. The bar is charged with non-­‐uniform charge density that can be described as λ(x) = c x2 , where c is some constant. Find the magnitude of the electric field at x=0. (a) cLk (b) (1/2) cLk (c) 2cLk (d) (1/3) cLk (e) 3cLk The electric field contribution is given by: dQ
dEx = k 2
x
2
L λ ( x ) dx
L cx dx
⇒ Ex = ∫ k
= ∫ k 2 =cLk
0
0
x2
x
14. An electron with a speed of 2×106 m/s moves into a uniform electric field of 500 N/C that is parallel to the electron's motion. How long does it take, in nanoseconds, for the electron to come to rest? (a) 23 (b) 35 (c) 12 (d) 1200 (e) never The velocity of the electron is parallel to the electric field, i.e. they are in the 


same direction. The electric force on the electron is F = −eE = ma , so it is decelerating. We can solve for the time it takes for velocity to become zero: eEt
v = v0 + at = v0 −
=0
m
v0 m
⇒t =
= 23ns
eE
Note that a vector has a specific direction, unlike a line which extends in two directions. So in physics “parallel” is always taken as in the same direction as another vector, and “anti-­‐parallel” the opposite direction. But those few students who thought parallel could also mean opposite were also given credit. 15. What is the flux leaving the surface of the shown cube if the electric field is given 
by E = −2xiˆ + 3yĵ and the cube has a side length of 2? (a) 8 (b) -­‐16 (c) 16 (d) -­‐8 (e) 0 Because of the directions of the electric field, the flux could only be non-­‐zero for the top-­‐bottom and left-­‐right faces. However, because the field also depends on x and y, the only non-­‐zero contributions are for the top and right faces: x
Phy 2049 Spring 2012 Exam 1 – v2 solutions Φ = Φ R + Φ T = E R ⋅ A R + ET ⋅ A T
= −(2)(2)(2)2 + (3)(2)(2)2 = 8
Recall that flux is positive when the field points out from a closed surface. 16. There are 4 nested shells of radii R=1, 2, 3, 4 m, all centered on the same point O. The shells, starting from the inner and moving outward, carry charges -­‐1, 2, -­‐3, 4 nC. Find the electric field magnitude in N/C at a distance r=2.5 m from the point O. (a) 1.4 (b) 2.9 (c) 4.3 (d) 5.8 (e) 14.4 The radius of 2.5m is between the second and third shells. Gauss’ Law tells us the field will be that due to the enclosed charges, which is the inner two shells. By symmetry they must have the same field as for a point charge: ( 2 − 1) 10 −9 C
9
2
2
E = 9 × 10 Nm / C
= 1.4 N/C ( 2.5 m )2
17. A uniformly charged thin non-­‐conducting shell (hollow sphere) of radius R with the total positive charge Q is placed at a distance d away from an infinite non-­‐
conducting sheet carrying a uniformly distributed positive charge with a density σ . The distance d is measured from shell’s center (point O). What is the magnitude of the total electric field at the center of the shell? R O d (
(a)
σ
2ε 0
)
(b)
Q
σ
+
4πε 0R 2 2ε 0
(
(c)
)
Q
σ
+
4πε 0R 2 ε 0
(d)
Q
4πε 0R 2
(e)
Q
σ
+
4πε 0R ε 0
By the superposition principle, the field at point O is the sum of the field contribution from the shell and the field from the sheet. But contribution to the field inside the shell from the shell itself is zero. Therefore the only contribution is from the non-­‐conduction sheet, which has a field contribution, independent of distance, of: Phy 2049 Spring 2012 E=
Exam 1 – v2 solutions σ
2ε 0
18. A charge Q is placed in the center of a shell of radius R. The flux of electric field through the shell surface is Φ0. What is the new flux through the shell surface, if its radius is doubled? (a) Φ0 (b) 2Φ0 (c) 4Φ0 (d) Φ0/2 (e) Φ0/4 The flux through any closed surface that encloses the charge must be the same, so Φ0 19. Four charges of the same sign and value q are placed in the corners of a square and free to move. One more charge Q is placed in the center of the square so that the entire system of the five charges has became statically stable, i.e. net forces on each of the five charges are equal to zero? Find the value of charge Q in terms of q. (a) -­‐0.96q (b) -­‐1.41q (b) -­‐1.12q (c) -­‐4q (d) -­‐0.71q q q Q= ? q q Let’s take for example the x-­‐component of the force on the lower left charge: q2
q2 2
Qq
2
Fx = −k 2 − k 2
+k
=0
2
d
2d 2
d/ 2 2
(
⇒ −1−
2 2Q 2
+
=0
4
2q
⎛
2⎞
⎜⎝ 1+ 4 ⎟⎠
⇒Q =
q = 0.96q
2
)
Phy 2049 Spring 2012 Exam 1 – v2 solutions 20 .Two very small spheres have equal masses m, carry charges of the same sign and value q, and hang on strings of length L as shown in figure. Due to the repulsive force, the spheres are separated by some distance d—find this distance. Assume that d<<L so that you can use the approximation tanα~sinα~α. α L d=? 3
(a) 2L
q2k
mg 3
(b) L
q2k
mg 2
Lq k
(d) 2mg 2
tan α =
FCoulomb
mg
⇒ d3 =
2kq 2 L
mg
q2
d
k 2
= d ≈ sin α = 2
mg
L 2
(b) 2L
q2k
mg 2
(c) L
q2k
mg