Download Exam 1 - UF Physics

Physics 2049
Exam 1
Spring 2002
q2
r12
q1
r13
q3
1. In the figure q1 = 300µC, q2 = 400µC, q3 = 500µC, r12 = 9m, r13 =
12m. Compute the magnitude of the total force on q3 .
F~3N ET = F~31 + F~32
kq3 q1
kq3 q2 =
cos
θ
î
−
sin
θ
ĵ
î
+
2
2
2
r13
r12
+ r13
r12
with θ = arctan( ) = 36.87◦
r13
~
F3N ET = 15.8N î − 4.8N ĵ
q
|F~3N ET | =
(15.8)2 + (4.8)2 N = 16.5N
1
q
Q
q
q
2. Three identical charges q are placed at the corners of the equilateral
triangle of side length L. The fourth charge Q is placed in the center
of the triangle. Find the value of Q in terms of q that would put the
whole system of four charges in equilibrium.
Calculating the net force on the top q and setting it to zero to enforce
equilibrium we get,
k|qQ|
2kq 2
F~qN ET = − 2 ĵ + 2 sin(60◦ )ĵ = 0
x
L
where x is the distance between q and Q. In terms of L,
x=
√
(L/2)
=
L/
3
cos(30◦ )
so that
√
3k|qQ|
3kq 2
=
L2
L2√
or Q = −q/ 3 (must have opposite sign for equilibrium)
e
E
2
(1)
3. An electron is placed in an electric field as shown. Find the direction
of the electrostatic force on the electron.
F = qE = −eE
Since F and E are antiparallel the force is directed to the left, ←.
4. A particle of mass 1 mg, and charge 6µC traveling with constant
velocity ~v0 = 10m/sî enters a region with a constant electric field
~ = 1.2N/Cĵ. What is the speed 2 seconds after it enters the region
E
with the electric field?
The velocity of the charge only changes in the ĵ direction,
F = qE = may ⇒ ay = qE/m = 7.2m/s2
vy (t) = ay t ⇒ vy (2s) = 14.4m/s
|~v (2s)| =
q
(10)2 + (14.4)2 )m/s = 17.53m/s
5. A spherical nonconducting balloon has a positive charge distributed
uniformly over its surface. If the balloon is expanded (no extra charge
is added) the magnitude of the electric field inside the balloon
, and
the magnitude of the electric field near the outer surface
. If a point
P is outside the balloon before expansion, and inside after expansion,
.
the magnitude of the E field at that point
Use Gauss’s law to get the electric field,
I
~ · d~a = E(4πr 2 ) = qin
E
0
S
kqin
E =
r2
Since all the charge is uniformly distributed on the surface of the balloon, the electric field inside is zero. Therefore, the field after expansion
will remain zero. A point just outside the surface of the balloon the
electric field will decrease since the electric field drops off as 1/r 2 . It
should also be evident that since the electric field is positive outside
the balloon, any point that starts outside and ends inside where the
field is zero will experience a decrease.
3
y
x
x=2cm
x=12cm
z
6. A cube with sides of length 10 cm rests on the y = 0 plane, with one
edge along the line x = 2cm and another edge along x = 12cm, as
shown in the above figure. A nonuniform field (in N/C) pierces the the
cube and is described by:
E(x) = Axî
where A =7.0E+6 N/(C·m). How much charge (in C) is inside the
cube?
Electric field flux only occurs at the x = 2cm and x = 12cm faces which
we will call left and right, respectively. Gauss’s law tells us,
I
~ · d~a = Q
E
S
Z0
=
right
~ · d~a +
E
Z
lef t
~ · d~a
E
The differential area elements on these surfaces are d~aright = dydz î and
d~alef t = −dydz î.
Q = 0 ×
E(x)
x=12cm
×L
2
− E(x)
x=2cm
where L is the side length. Therefore,
Q = 0 × 7000Nm2 /C = 6.2E − 8C
4
×L2
7. The electric field at the surface of a sphere of radius R is measured.
It is found to be radially outward and to have the same magnitude all
over the sphere. Call that magnitude E1 . A second sphere of radius
2R is found to have a radially inward electric field that is uniform also
but with magnitue 4E1 . What is the ratio of the charge in the second
sphere to the charge in the first sphere (Q2 /Q1 )?
If we place Gaussian surfaces on the surfaces of both spheres they would
tell us the magnitudes of the electric field are,
k|Q2 |
k|Q1 |
, E2 =
2
R1
R22
R22 |Q1 |
1
4|Q1 |
=
= =
2
R1 |Q2 |
4
|Q2 |
E1 =
−→
E1
E2
Since the fields point in opposite directions, the signs of Q1 and Q2 are
opposite and we have
Q2
= −16
Q1
8. A clever physicist produces a very long, cylindrical distribution of electrical charge. The diameter of the cyclindrical distribution is D. Inside
the volume charge density varies radially as,
ρ(r) = ρ0 1 −
2r
D
For r > D/2, ρ(r) = 0. Here r is the radial distance from the cylindrical
axis and ρ0 is a negative constant ρ0 < 0. Calculate the electric field
generated by rho for all r > D/2. That is, at any point outside of the
region containing the charge.
Again we use Gauss’s law, but this time for cylindrical symmetry,
I
S
~ · d~a = E(2πrL) = qin /0
E
qin =
Z
dV ρ(V ) =
= 2πL|ρ0 |
Z
D/2
0
5
Z
2π
0
dφ
Z
0
L
dz
2r
dr r 1 −
D
Z
D/2
0
dr rρ(r)
D2 D2
−
= 2πL|ρ0 |
8
12
D2
= πL|ρ0 |
12
2
|ρ
|D
0
~ =
−→ E
inward
240 r
Now by the way of the 5 possible answers, only 2 had the electric field
directed inward, allowing you to eliminate 3 of the possible choices. Of
the remaining two, only the one derived above has the correct dimensions of of an electric field. Therefore, one could have answered this
question without having to even know what an integral is!
!
9. The electric field in megaN/C at a point halfway between a point charge
of +2µC at the origin and a point charge of -3µC at x = +10 cm is:
Between the two charges, the electric field is pointed to the right and
the components add,
k|q2 |
k|q1 |
+
E(x) =
2
x
(0.1m − x)2
k(2E-6C+ 3E-6C)
E(0.05m) =
= 1.8E7 N/C = 18 megaN/C
(0.05m)2
10. An electric dipole is composed of equal and opposite charges each of
magnitude 0.1µC, placed at y = −1 cm, respectively. The magnitude of
the electrostatic field due to this dipole (in N/C) at the point x = +3m
is:
The magnitude of the electric field along the z axis due to the two
charges along the y axis comes from adding the y components and
cancelling the z components to give,
kqd
2kq
sin θ =
E(z) =
2
2
(d/2) + z
((d/2)2 + z 2 )3/2
where sin θ = √
d/2
(d/2)2 +z 2
, and d is the distance between the charges.
The dipole field comes from approximating the electric field for z d,
so we can neglect the d/2 in the denominator and we are left with,
E(z) ≈
kqd
(1E-7C)(0.02m)
=
k
= 0.67 N/C
z3
(3m)2
6