Download 2011 B R = 0.12 m). Therefore,

2011 B
2. (a) i. Since copper is a conductor, all of the charge is located at the surface (R1 = 0.12 m). Therefore,
the electric field everywhere inside the copper sphere is zero. So, the potential inside this sphere
is a constant (everywhere the same) and equal to the potential at a point just outside the surface
of the sphere. Assume zero potential at infinity.
6.4 x 10 -9 C
Q
V = k = 9 x 10 9 N × m 2 / C 2
0.12 m
r
(
)
V = 480 V
ii. V = k
6.4 x 10 -9 C
Q
= 9 x 10 9 N × m 2 / C 2
0.24 m
r
(
)
V = 240 V
Volts
(b)
V
240
r
R1
(c) i. Zero as described above
ii. E = k
-9
Q
C
9
2
2 6.4 x 10
=
9
x
10
N
×
m
/
C
2
2
r
(0.24 m)
(
)
E = 240 N / C
(d)
R1
R2
V
O
x
2010 B
3. (a)
q2 ____ Negative
X Positive
____
q1 ____
X Negative
____ Positive
(b)
F2
F1
(c) F = å Fx = -F1 x -F2 x = -k
(
= - 9.0 x 10 9 N × m 2 / C 2
)
q1 q3
-k
q2 q3
r2- 3
r1- 3
é 4.0 x 10 -6 C 1.0 x 10 -6 C
ù
1.7. x 10 -6 C 1.0 x 10 -6 C
cos
37
+
cos
53
ê
ú
( 4.0 m) 2
(3.0 m) 2
êë
úû
2
(
2
)(
)
(
)(
)
F = - 2.8 x 10 -3 N
(d) F = - 2.8 x 10 -3 N
- 2.8 x 10 -3 N = 1.0 x 10 -6 C E
(
)
E = - 2.8 x 10 3 N / C
E = å E x = -E1 x -E 2 x = -k
(
= - 9.0 x 10 N × m / C
9
2
2
)
q1
r1- 3
(
2
OR (The long method)
q
- k 22
r2- 3
(
)
)
é 4.0 x 10 -6 C
ù
1.7. x 10 -6 C
cos
37
+
cos
53
ê
ú
2
(3.0 m) 2
êë ( 4.0 m)
úû
E = - 2.8 x 10 3 N / C
(e)
X
The net force on charge q3 is currently to the left, so this additional charge would have to create a force
equal and opposite to the current net force to produce a new net force of zero. Therefore, this additional
force would have to be to the right. Since, q3 and this new charge are both positively charged, and like
charges repel, placing this new charge to the left of charge q3 would create a force to the right.
2009 B
2. (a)
1
(b) V A = V1 A + V 2 A = k
VA =
kQ
kQ
+
L sin q L sin q
VA =
2kQ
L sinq
Q
Q
+k
rA
rB
Note: rA = rB = L sinq
(c)
FE2
FT
1
Fg
(d) SFx = FTx - FE 2 = 0
SFy = FTy - Fg = 0
2
2006 B
3. (a)
X Positive
____
____ Negative
Since q1 is negative, the electric field due to this charge at point P is to the right. If the net electric
field is to be zero at point P, then the field from charge q2 must be equal but OPPOSITE. A
positive charge for q2 would produce an electric field that is to the left at point P.
(b) SE = E1 - E 2 = 0
q
q
k 1 =k 2
r2
r1
-9
q2
3.0 x 10 C
=
0.10 m
0.30 m
q2 = 9 x 10 -9 C
(c) E 21
-9
q1
C
9
2
2 -3.0 x 10
=k
= 9.0 x 10 N × m / C
r1
0.30 m
(
)
E21 = -90 N/C
(d) V NET = V 2 - V1 = 0
Note: r1 + r2 = 030
. m, so r1 = 030
. m - r2
q
q2
=k 1
r1
r2
-9
9.0 x 10 C 3.0 x 10 -9 C
=
r2
0.30 m - r2
r2 = -0.225 m. So, zero potential would occur 0.225 m to the left of q2 which would be at position
k
-0.05 m on the x-axis.
(e) Zero work because W = qDV and the potential at an infinite distance away is zero, so DV = 0.
Therefore, the work done in bringing an electron from infinity to this zero-potential point would
be zero.
2005 B
3. (a) E NET = E 2 - E1 = k
E NET
Q
Q2
+2q
+q
- k 21 = k 2 - k 2
2
r
r
a
a
O
q2
= k 2
a
E1
(b) VNET = V1 + V2 = k
VNET
E2
Q
Q1
+q
+2q
+ k
+ k 2 = k
r2
r1
a
a
3q 2
= k
a
(c) i. F31 = k
Q1Q 3
r31
F31 = - k
ii. F32 = k
2
(
( + q)(- q)
a2 + x0
2
)
2
q2
2
a2 + x0
Q 2Q 3
r32
F32 = - k
= k
2
= k
(
( +2q)(- q)
a2 + x0
2
)
2
2q 2
2
a2 + x0
(d)
a +q
A
B
-a +2q
C
2001 B
3. (a) i. Let d be the length of the diagonal of this square.
s2 + s 2 = d2
2s2 = d2
d= 2s
let r = the distance from any of the corners of the square to the center (these are equidistance).
2
r = ½d =
s
2
q
q
q
q
+Q
-Q
+Q
-Q
Vtotal = V1 + V2 + V3 + V4 = k 1 + k 2 + k 3 + k 4 = k
+k
+k
+k
r
r
r
r
r
r
r
r
Vtotal = 0 V
ii. Etotal = 0 N/C since the x-components will add to zero and the y-components will add to zero due to the
symmetrical arrangement of the charges.
q
q
q
q
+Q
+Q
-Q
-Q
+k
+k
+k
(b) i. Vtotal = V1 + V2 + V3 + V4 k 1 + k 2 + k 3 + k 4 = k
r
r
r
r
r
r
r
r
Vtotal = 0 V
ii. The y-components of the total electric field will add to zero due to the symmetrical arrangement of the
charges. However, all of the x-components will be equal in magnitude and have direction to the right.
The electric field is always away from a positive charge and toward the negative charges.
Etotal = Etotalx = E1x + E2x + E3x + E4x = E1x + E1x + E1x + E1x = 4E1x, since E1x = E2x = E3x =E4x
Q
Q
Q
Etotal = 4k 2 = 4k
= 4k
2
2 2
r
æ 2 ö
s
ç
÷
s
4
ç 2 ÷
è
ø
Etotal = 8k
Q
s2
X Arrangement 1
(c) ___
Calculations for the total potential difference for Arrangement 1:
q
q
q
+Q
-Q
+Q
2Q
Q
2Q
DV1 =Vtota1l - 0 = V1 + V2 + V3 = k 1 + k 2 + k 3 = k
+k
+k
=k
-k
+k
s
s
s
s
s
2s
2s
2s
2s
(2 2 - 1)Q
Q
DV1 = k
= 129
. k
s
2s
q
q
q
+Q
-Q
+Q
+k
+k
DV2 = Vtota12 - 0 = k 1 + k 2 + k 3 = k
s
s
s
s
s
2s
Q
DV2 = k
2s
This shows that DV1 > DV2 . Since W = qDV, arrangement 1 will require more work to remove the
particle at the upper right corner from its present position to a distance a long way away from the
arrangement.
2000 B
7. (a)
__ Positive
X
__Negative
__ Neutral
__ It cannot be dtermined from this information.
The right-hand rule (put your fingers in the direction of the moving charged particles, rotate your
wrist until your fingers can curl in the direction of the magnetic field, and your thumb will indicate
the direction of the magnetic force on the moving charges, thus, the direction of the center of the
circular path followed by these moving charges) is for positive charges. Applying the right-hand
rule to these charges gives a direction toward the center of the circular path that is toward the top
of the page which is opposite that shown in the diagram. [or the left-hand rule would give the
results in the diagram].
(b)
(c) In the region between the plates both a magnetic force and electrostatic force are acting on the
moving charges.
SF = Fm - Fe = ma = 0
qvB = qE
E = vB = (1.96 x106 m/s)(0.20 T)
E = 3.8 x 105 V/m
V = Ed = (3.8 x 105 V/m)(6.0 x 10-3 m)
V = 2.28 x 103 V = 2280 V
(d) In the region outside the plates only a magnetic force is acting on the moving charges resulting in
circular motion.
SF = Fm = Fc
Units Analysis:
v2
qvB = m
R
C
m
q
v
19
. x10 6 m / s
1
1
A· m
C
s
=
=
=
=
=
=
m BR (020
. T )(010
. m)
m
Ts ( N / A · M ) · s N · s
kg
( kg 2 )s
s
q
7
= 9.5 x 10 C/kg
m