Midterm 1 Solutions. Professors: Korytov, Takano. PHY 2049, Spring 2014.
February 10, 2014. 1) The force between the proton and electron is as follows. F
= k.
Midterm 1 Solutions Professors: Korytov, Takano PHY 2049, Spring 2014 February 10, 2014 1) The force between the proton and electron is as follows F =k
e2 r2
(1)
Next, make use of F = ma to find the acceleration 2
F e2 (1.6 × 10−19 ) a= =k 2 = 9 × 109 = 9.0 × 1022 m/s2 me r me (5.3 × 10−11 )2 (9.1 × 10−31 ) (2)
2) The force on q3 from q1 is as follows F13 = k
q1 q3 2 r13
(3)
This can be split into x and y components F13x = k
q1 q3 , 2 r13
F13y = 0
(4)
Similarly F23 is the force on q3 from q2 F23 = k
1
q 2 q3 2 r23
(5)
Splitting this force into x and y components is a little harder. The x and y components of F23 can be written F23x = k
q2 q3 cos θ, 2 r23
F23y = k
q2 q3 sin θ 2 r23
(6)
Where 2 2 2 r23 = r13 + r12
(7)
Since the angle is not given, it can be found by writing it in terms of the side lengths of the triangle. cos θ = r13 /r23 (8) sin θ = r12 /r23 Thus we can rewrite the x and y components of F23 q2 q3 r13 q2 q3 r12 F23x = k 2 , F23y = k 2 r23 r23 r23 r23
(9)
Combining the x components and the y components we get Fx = F13x + F23x ,
Fy = F13y + F23y
Finally the magnitude of the force on q3 is obtained q F = Fx2 + Fy2 = 16.5N
(10)
(11)
3) For the third charge to be in equilibrium (no force on it), it must be placed where the electric field from q1 and q2 is zero. This will be somewhere between q1 and q2 as they are both positive charges. Consider a point a distance d from q1 . The magnitudes of the electric fields from each of the first two charges will be as follows
2
Eq = k dq2 (12) 4q E4q = k (L−d) 2
At the spot where the electric field is zero, these will be equal in magnitude. q 4q 2 = 2 d (L − d)
(13)
Cross multiplying and solving for d results in d = L/3. The force on q must also be zero for a system in equilibrium. Fq = FqQ + Fq4q = k
q4q qQ +k 2 =0 2 d L
(14)
So −
qQ 4q 2 = L2 /9 L2
(15)
Solving for Q/q, Q/q = −
4 9
(16)
From Newtons third law it is known that because the net force on the other two charges is zero, the net force on the third will also be zero. However, it can be seen explicitly
F4q = F4qq + F4qQ
4q 2 4qQ 4kq 2 =k 2 −k = − L L2 (2/3L)2
9 4qQ =0 4 L2
(17)
4) The electric field a distance r from a dipole of moment p = qd along it’s axis is E = 2πεp0 r3 . Examine the force on each of the charges within a 3
second dipole placed at a distance x from the first. F−q = −qE (r = x − d/2) =
−qp −qp 1 1 (18) 3 = 3 2πε0 (x − d/2) 2πε0 x (1 − d/2x)3
Using the given approximation it is found that −qp 3d F−q = 1+ 2πε0 x3 2x
(19)
Using a similar process for the positive charge in the second dipole qp 3d F+q = qE (r = x + d/2) = (20) 1− 2πε0 x3 2x Add these two together to get the total force qp 3d 3d qp 6d Ftot = F−q + F+q = +1− −1 − =− 3 2πε0 x 2x 2x 2πε0 x3 2x
(21)
Using p = qd Ftot
−3p2 = 2πε0 x4
(22)
5) The forces must add to make an angle of θ. Thus tan θ = Solving for E E=
qE mg
mg tan θ = 120 N/C q
(23)
(24)
6) Start by finding the acceleration for each particle. The force on each is eE ap = eE/mp , ae = eE/me (25) 4
If the distance of the collision from the positive plate is x, then the distance that each particle travels is 1 x = ap t2 , 2
1 L − x = ae t2 2
(26)
Solving for t2 and setting the two equations equal to each other gives 2xmp = 2 (L − x) me
(27)
solving for x gives x=L
me me + mp
= 0.0005L
(28)
7) Options 3 and 4 give electric fields of zero at the center because charges opposite each other are of like charge. Option 2 has two charges that cancel, if it is assumed the side of the square has length d, the other two combine to give q2 q2 (29) E = 2k √ 2 = k 2 d 2d Option 1 has four charges that contribute to the electric field with the horizontal components canceling. Each charge contributes the following to the electric field in the vertical direction. √ ! 2 q2 2 q k √ 2 cos (45) = k 2 (30) 4 d 2d Combining all four charges gives Ey =
√
5
2k
q2 d2
(31)
Because this is larger than any of the other options, this is the correct answer.
8) The electric field E from a thin sheet of charge with surface charge density σ is E = 2εσ0 . Therefore at point A, the sheets provide electric fields as follows E (+σ) = σ/2ε0 → E (−2σ) = 2σ/2ε0 ← E (+3σ) = 3σ/2ε0 ←
(32)
Adding these together it is found 5σ σ 2σ − = 2ε0 2ε0 ε0
(33)
9) The flux through the top, bottom, front and back faces of the cube will all be zero because the electric field is parallel to those faces. The flux through the left side of the cube will be Z Z Z 6 N ~ (0.02m) 0.01m2 Φlef t = E· d~a = − Axda = −Ax da = − 7 × 10 C· m (34) Similarly, the flux through the right side of the cube will be Z Z Z 6 N ~ Φright = E· d~a = Axda = Ax da = 7 × 10 (0.12m) 0.01m2 C· m (35) This gives a total flux of Φ = Φlef t + Φright = −1400
N· m N· m N· m + 8400 = 7000 C C C
(36)
Use gauss’s law to find the charge inside qenc = ε0 Φ = 6.2 × 10−8 C 6
(37)
10) A spherical Gaussian surface at 1.5R will have zero flux through it because the electric field is zero inside a conductor. By Gauss’s law, qenc = ε0 Φ, there is no net charge inside this sphere. Therefore, there must be a charge of +q on the inside surface of the shell. Since the shell has a net charge of +2q, the outside of the shell has +q on it.
11) See the solution to the previous problem.
12) To find the electric field at the surface, use Gauss’s law. I I I ~ d~a = Eda = E da = E4πR2 Φ = E·
qenc 4 πR3 ρ 3
= ε0 Φ = ε0 E4πR2
(38)
(39)
Solving for E gives E=
Rρ 3ε0
(40)
So where the electric field is E = Rρ/27ε0 the E field will be down by a factor of 9 from the value at the surface. Inside the sphere this implies the following. E ace A ρV = ε0 surf 9 (41) Rρ 4 3 2 ρ 3 πr = ε0 27ε0 4πr Where solving for r gives r = R/9. There will also be a value of r outside the sphere which can be found as follows 4 Rρ ρ πR3 = ε0 4πr2 3 27ε0 This gives r = 3R. 7
(42)
13) Because the difference in electric potential between two points is independent of the path taken, find ∆V by using Z ~ d~s ∆V = Vf − Vi = − E· (43) and integrating perpendicular to the E field, and then along an equipotential line which wont contribute to the change in potential. The electric field points directly down towards the plate, so integrate directly up from point O to the height of point P, then along the horizontal equipotential line to point P. Z Z Z ~ ∆V = − E· d~s + 0 = Eds = E ds = Ed (44) The electric field above a solid plate with surface charge σ is E = σ/ε0 , thus ∆V = σd/ε0 .
14) The electric field is the gradient of the electric potential. ~ = ∂V ˆi + ∂V ˆj + ∂V kˆ = 3x2ˆi + z 2ˆj + 2zy kˆ E ∂x ∂y ∂z
(45)
For point (x, y, z) = (1, 1, 1), this gives ~ = 3ˆi + 1ˆj + 2kˆ E The magnitude of this vector is √ N E = 32 + 12 + 22 = 3.7 C
(46)
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15) The electric potential at the surface of a sphere of charge q is V = kq/r. The electric potential at the surfaces of the spheres are thus VA = k
2σ 4πrA A rA
(48) VB =
4πr2 σ k rBB B
8
Any two points connected by a conducting wire must be at the same electric potential, therefore VA = VB . Setting the previous equations equal to each other and solving for σA /σB , the following is obtained σA /σB = rB /rA = 1/2
(49)
W = ∆U = q∆V = q3 (VB − VA )
(50)
VB = kq1 /a + kq2 /b = 0V VA = kq1 /b + kq2 /a = −1.2 × 1012 V
(51)
16)
Plugging numbers into the first equation, we get W = 3 × 1012 Joules.
17) First find the charge on each capacitor before the switches are closed. q1 = C1 (100V) = 100µC q2 = C2 (100V) = 300µC
(52)
Next, consider what happens when the switch is closed. Because the two capacitors are oppositely charged, there is −300µC on the top plate of C2 , and +100µC on the top plate of q1 . These two plates are connected when the switch is closed, therefore they will share −200µC of charge between them. Likewise, the bottom plates of each capacitor will share +200µC. q = CV
(53)
Also, after the switches are closed, the voltage drop across each capacitor must be equal, V1 = V2 . Therefore C1 1 q1 = q2 = q2 (54) C2 3 9
This means that q1 = 50µC and q2 = 150µC. Solving either capacitor for the voltage, using q = CV , gives a voltage drop of 50V.
18) Adding the two sets of capacitors arranged in parallel gives an equivalent capacitance of 2C for each. This leaves three capacitors in series, two of which have capacitance of 2C, the third has capacitance C. 1 1 1 4 1 = + + = Ceq 2C C 2C 2C
(55)
Inverting this we get Ceq = C/2.
19) C1 and C2 are in parallel, so their equivalent capacitance is 3µF. There are now two capacitors in series, each with capacitance 3µF. There are 6V across these two capacitors. Because they are equal, the voltage splits evenly between them. Therefore there is a 3V drop across C1 . q1 = C1 V1 = (1µF) (3V) = 3µC
(56)
20) Treat the sphere as one half of a capacitor, the other half being at R = ∞. The capacitance of an isolated sphere of radius R is C = 4πε0 R. The energy stored in the electric field of a capacitor is U = q 2 /2C. Combining these two for a sphere with charge q = Q, we get U=
Q2 8πε0 R
10
(57)
