Download YEAR 12 PHYSICS ELECTROSTATICS REVISION SHEET 2

YEAR 12 PHYSICS ELECTROSTATICS REVISION SHEET 2
Question 1
Draw a simple circuit with one resistor and a cell. Place a voltmeter and an ammeter in correct positions.
V
A
..
Question 2
Draw a simple E vs r graph showing the relationship of electric field strength as distance (r) is increased.
E(NC-1)
r(m)
Question 3
A 1.2 m length of copper wire has a cross-sectional area of 1.72  10-5 m2. What is its resistance?

RA
L
ρCu= 1.7 × 10-8 = R × 1.72 × 10-5 /1.2
∴ R = 1.186mΩ
Base your answers to questions 4 through to 7 on the circuit diagram below.
Question 4
What is the potential difference across R2?
24V
Question 5
What is the current through R1?
2A
Question 6
What is the overall equivalent resistance of
the circuit?
3Ω
Question 7
What is the current through the ammeter?
8A
Question 8
After two neutral solids, A and B, were rubbed
together, solid A acquired a net positive charge. What net loss or
gain of electrons would solid B undergo?
Net gain of electrons
Question 9
How many electrons are lost to produce a charge of +2.0 × 10-19 C ?
≈ 1 electron
Question 10
(a.) Calculate the work required to move an alpha particle ( a Helium nucleus ) from the negative
to the positive terminal of a24V battery. Convert this answer to eV.
W = qV = 2 × 1.602 × 10-19 × 24 ≈ 7.7 × 10-18J = 48eV
(b.) What would be the velocity of the particle as it hit the negative terminal?
K.E. = qV
½ × 4 × 1.6606 × 10-27kg × v2 = 7.7 × 10-18
∴ v = 4.8 × 104 ms-1
Question 11
A charge of -4µC is placed at 30cm on the positive x-axis and a charge of +6µC is placed at
70 cm on the positive x-axis. Calculate the electric field strength and direction at x = 90cm.
E4
• -4.0μC
E = E6 – E4
• +6μC
E6
• +1C
0.3m
0.7m
0.9m
k  6 C k  4 C

=
= 1.25 × 106 NC-1
0.2 2
0.6 2
Question 12
In a Millikan oil drop experiment an oil drop of mass 2.08  10-15 kg was held stationary between a pair of electric plates
held 5.0 cm apart. The electric field strength between the plates is set to 5500 Vm-1.
E= 5500 Vm-1
(a.) How much electric charge was present on the oil drop?
F = Eq = mg
q = mg/E = 2.08 × 10-15 × 9.8/5500 ≈ 3.7 × 10-18C
(b.) How many excess electrons does this correspond to?
3.7 × 10-18/1.602 × 10-19 ≈ 23 electrons
(c.) What was the potential difference between the plates?
V = Ed = 275V
Question 13
Calculate the absolute potential at point B under the influence of the charges shown.
 +7µC
0.3m
B
0.9m
● +9 μC
0.6m
 -8µC
V = kq/r
V = k × 7.0 × 10-6/0.3 + k × 9 × 10-6/0.9 + k × -8 × 10-6/0.6 = 180 000V
Question 14 Two charged particles are fixed in place. One with charge Q1 is placed at 15cm on the x-axis and the
other of charge Q2 = -2Q1 is placed at 60cm on the x-axis. A third charge, Q3 is placed on the x-axis so that they are all at
equilibrium.
Calculate where Q3 must be placed.
F2,3
F1,3
Q3
Q1
Q2
0.15m
0.60m
x
Q3 must be placed to the left of Q1, regardless of whether it’s negative or positive as it would not be at
equilibrium in any other location. (on the test you would need to elucidate on this.)
Fnet = 0 = F1,3 – F2,3 = k × Q1Q3/(x – 0.45)2 – kQ2Q3/x2
Q1x2 = Q2(x – 0.45)2
Q1x2 = 2Q1(x – 0.45)2 ******DO NOT INCLUDE THE
NEGATIVE SIGN AS IT HAS ALREADY BEEN ACCOUNTED FOR IN THE VECTOR DIAGRAM
x2 = 2(x – 0.45)2
x2 = 2x2 - 1.8x + 0.405
x2 – 1.8x + 0.405 = 0
∴ x = 1.54m, 0.264m
x is the distance from Q2, so 0.264m is impossible as it would lie between Q1 and Q2. The actual answer is
1.54m behind 60cm, so it would be expressed as -0.94m
Question 15
Current ( Amperes )
The graph below gives the current versus potential difference of two resistors X and Y.
5
Y
4
X
3
2
1
0
2
4
6
8
10
12
Voltage ( Volts )
X and Y are connected in series and a potential difference is applied across the combination, so that a current of 2.0A
flows in resistor X.
(a.) What is the potential difference across resistor X?
≈ 5.9V
(b.) What is the current through resistor Y?
2.0A
(c.) What is the potential difference across resistor Y?
≈ 4V
(d.) What is the potential difference across the combination of resistors?
9.9V
(e.) Calculate the resistance in Ohms of each resistor
The inverse of the slope represents R.
slopeX = (5.9 – 0)/(2 – 0) = 2.95Ω
slopeY = (8 – 0)/(4 – 0) = 2Ω
Question 16
A student conducts an experiment in order to investigate the effect two charges have on each other at various distances.
The set-up used is illustrated below. Here a +3.0µC charge is fixed to a table and is well insulated. A very light ball
with an unknown charge is attached to a spring balance through an insulator.
gh
The unknown charge q is moved to
various distances d from the +3.0µC
stationary charge and the force
required to hold the ball in
equilibrium at the desired distance is
measured using a spring balance.
The following results were obtained
:
(a.) Would the unknown charge q carry excess electrons or is it electron deficient?
It would carry excess electrons, as it is stretching the Newton balance towards
the positively charged sphere.
Distance d
(cm)
5
10
15
20
25
30
Force F (N)
25.9
6.5
2.9
1.6
1.0
0.7
(b.) By considering the general proportionality relationship between Force and distance, and using your knowledge of
Coulomb’s Law, redraw the data table including a column that will help you draw a linear graph. Use this graph to
determine the value of the unknown charge q.
Force Vs Distance
0.05
0.10
0.15
0.20
0.25
0.30
Force(N)
d-2 (m-2)
25.9
6.5
2.9
1.6
1
0.7
30
25
400.00
100.00
Force(N)
Distance(m)
44.44
25.00
20
F = 0.0618d-2
15
10
5
16.00
0
0.00
11.11
0.05
0.10
0.15
0.20
Distance(m)
Force Vs Inverse Square of Distance
30
25
slope = 0.0648Nm2 = kq1q2
F = 0.0648d-2
Force(N)
20
15
q1 = -2.4 × 10-6C
10
5
0
0.00
100.00
200.00
d-2(m -2)
300.00
400.00
500.00
0.25
0.30
0.35
(c.) Draw a diagram to illustrate the electric field that exists on the surface of the workbench when q is 20 cm from the
+3.0µC charge and calculate the size of this field at a point halfway
E3
E-2.4
+3μC
-2.4μC
0m
E = E3 + E-2.4
E = 4.86 × 106NC-1 t the right of the table.
0.1m
0.2m
Question 17
Two pith balls shown below have a mass of 10.0g each and have equal charges. One pith ball is suspended by an
insulating thread. The other is brought to 2.0 cm from the suspended ball. The suspended ball is now hanging with the
thread forming an angle of 20.00 with the vertical. The ball is at equilibrium. Calculate the value of the charge on each
ball.
Tsin200
20o
Tcos200
T
Fq
2.0cm
Horiz
Fx = Fq – Tsin200 = 0
mg
kq2/0.022 = Tsin200
Vert
Fy = 0 = Tcos200 - mg
∴ T = 0.01 × 9.8/cos200 = 0.1043N
∴ q2 = 0.1043sin200 × 0.022/(9 × 109)
q = 3.9 × 10-8C
Question 18
An electron moving with a velocity of 3.0 x 107ms-1 enters a uniform electric field of strength
2.0  104NC-1 as shown in the figure below.
E = 2.0  104 NC-1
vx
v
e
vy
v = 3.0  107 ms-1
Calculate the velocity of the electron after 2.5 x 10-8s in the electric field.
Horiz
ux = 0 F = ma = Eq a = (2.0 × 104 × 1.602 × 10-19)/(9.11 × 10-31) = 3.52 × 1015ms-2, t = 2.5 × 10-8s
vx = ux + axt = 8.8 × 107ms-1 (to the left, due to E direction)
Vert
uy = vy = 3.0 × 107ms-1 ∴ v ≈ 9.3 × 107ms-1 , ϕ = tan-1((8.8 × 107)/(3.0 × 107) = 71.20 ∴ θ = 161.20