electrostatics - grey college
TRANSCRIPT
Electrostaticsp. 213
Negative Charge - Addition (gain) of e-
+ +
+ +-
-
-
-
+ +
+ +-
-
-
-
+ +
+ +-
--
-
-
-
-
-
Positive Charge - Removal (loss) of e-
+ +
+ +-
-
+ +
+ +-
-+ +
+ +-
- ---
-
Similar charges repel each other.
-- +
-- -- --
Unlike charges attract each other.
+++
-- --- -
Law of Conservation of Charge:
Charge cannot be created or destroyed,
it can only be transferred from one
object to another.
Coulomb’s Law:
The magnitude of the electrostatic force
exerted by two point charges (Q₁ and Q₂)
on each other is directly proportional to the
product of the magnitudes of the charges
and inversely proportional to the square of
the distance (r) between them.
2
21
r
QkQ F =
F – Electrostatic force in Newton
Q1 and Q2 – Charges in Coulomb
r – Distance between charges in m
k – proportionality constant
(9 x 109 N.m2.C-2)
Coulomb’s Law
(Electrostatic forces between charges)
Two factors influence
the force:
• The magnitude of the
charges.
• The magnitude of the
distance between the
charges.
21QQ α F
2r
1 α F
ExampleThe centres of two identical spheres are 6 mm
apart. They carry charges of 5 nC and -10 nC
respectively as shown in the diagram.
1. Calculate the magnitude of the electrostatic
force between the two spheres.
Q1 Q2
5 nC
-10 nC
6mm
F = 𝑘𝑄1𝑄2
𝑟2
= (9×109)(5×10−9)(10×10−9)
(0,006)2
= 1,25 × 10-2 N, attractive
It is not necessary to substitute the signs of the charges.
Example
The centres of two identical spheres are 6 mm
apart. They carry charges of 5 nC and -10 nC
respectively as shown in the diagram.
2. Is it a force of attraction or repulsion?
Q1 Q2
5 nC
-10 nC
6mm
Attraction
Example
The centres of two identical spheres are 6 mm
apart. They carry charges of 5 nC and -10 nC
respectively as shown in the diagram.
3. The spheres touch each other, and are then
moved back to their original positions. What
charge does each sphere have now?
Q1 Q2
6mm
Qon each after separation = 𝑄1+𝑄2
2
= +5×10−9 + (−10×10−9)
2
= -2,5 x 10-9 C
Example
The centres of two identical spheres are 6 mm
apart. They carry charges of 5 nC and -10 nC
respectively as shown in the diagram.
4. How many electrons are transferred when
they touch, and in which direction?
Q1 Q2
6mm
ΔQ1 = Q1new - Q1original
= (-2,5x10-9) - (+5x10-9)
= -7,5 x 10-9 C
Thus, number of electrons transferred:
ne- = 7,5×10−9
1,6×10−19
= 4,7 x 1010 electrons, from Q₂ to Q₁
Homework
p. 224,
nos. 3, 6.1, 6.2, 11, 13, 14
Electrons are removed
Forces in two dimensions:
BC
A -
++F
A o
n C
FB on C Cθ
Additional examples:1. For the charge configuration shown, calculate the charge on Q₃ if the resultant force on
Q₂ is 6,3 x 10-5 N to the right, and Q1 = 4,36 x 10-6 C, Q2 = -7 x 10-7 C, r1 = 1,85 x 10-1 m,
r2 = 4,7 x 10-2 m.
Q3Q1 Q2
r1 r2
FQ1 on Q2 = k𝑄1𝑄2
𝑟2 = 9 x 109 x (4,36×10−6)×(7×10−7)
(1,85×10−1 + 4,7×10−2)2
= 0,51 N, to the left
+ ? -
Fnet = FQ1 on Q2 + FQ3 on Q2
6,3 x 10-5 = -0,51 + FQ3 on Q2
FQ3 on Q2 = 5,1 x 10-1 N, to the right
Additional examples:1. For the charge configuration shown, calculate the charge on Q₃ if the resultant force on
Q₂ is 6,3 x 10-5 N to the right, and Q₁ = 4,36 x 10-6 C, Q₂ = -7 x 10-7 C, r₁ = 1,85 x 10-1 m,
r₂ = 4,7 x 10-2 m.
FQ3 on Q2 = k𝑄3𝑄2
𝑟2 = 9 x 109 x ( 𝑄3 )×(7×10−7)
(4,7×10−2)2
5,1 x 10-1 = 2 851 969,217 x Q3
∴ Q3 = -1,79 x 10-7 C
Q3Q1 Q2
r1 r2
+ - -
Additional examples:2. Calculate the resultant force on Q₁ given the following charge configuration:
Q2Q1
Q3
0,04m
+3nC
FQ3 on Q1 = k𝑄1𝑄2
𝑟2 = 9 x 109 x (3×10−9)×(2×10−9)
(0,04)2
= 3,375 x 10-5 N, downwards
0,07m
+2nC
+1nC
Additional examples:2. Calculate the resultant force on Q₁ given the following charge configuration:
Q2Q1
Q3
0,04m
+3nC
FQ2 on Q1 = k𝑄1𝑄2
𝑟2 = 9 x 109 x (1×10−9)×(2×10−9)
(0,07)2
= 3,673 x 10-6 N, to the left
0,07m
+2nC
+1nC
Additional examples:2. Calculate the resultant force on Q₁ given the following charge configuration:
Q1
3,67 x 10-6 N
3,38 x 10-5 N
3,38 x 10-5 N
θ
(Fnet)2 = (3,67 x 10-6)2 + (3,38 x 10-5)2
Fnet = 3,4 x 10-5 N
tan θ = (3,38 x 10-5) ÷ (3,67 x 10-6)
θ = 83,8o
∴ Fnet = 3,4 x 10-5 N; 186,2o from the line connecting Q₃ with Q₁
Additional examples:3. Calculate the resultant force on Q₁ given the following charge configuration:
Q2Q1
Q3
0,4m
-3nC
FQ3 on Q1 = k𝑄1𝑄2
𝑟2 = 9 x 109 x (3×10−9)×(9×10−9)
(0,4)2
= 1,52 x 10-6 N, downwards
0,65m
-9nC
+1nC
Additional examples:3. Calculate the resultant force on Q₁ given the following charge configuration:
Q2Q1
Q3
0,4m
-3nC
FQ2 on Q1 = k𝑄1𝑄2
𝑟2 = 9 x 109 x (1×10−9)×(9×10−9)
(0,65)2
= 1,91 x 10-7 N, to the right
0,65m
-9nC
+1nC
Additional examples:3. Calculate the resultant force on Q₁ given the following charge configuration:
Q11,91 x 10-7 N
1,52 x 10-6 N
1,52 x 10-6 N
θ
(Fnet)2 = (1,52 x 10-6)2 + (1,91 x 10-7)2
Fnet = 1,53 x 10-6 N
tan θ = (1,52 x 10-6) ÷ (1,91 x 10-7)
θ = 82,8o
∴ Fnet = 1,53 x 10-6 N; 172,8o from the line connecting Q₃ with Q₁
Additional examples:4. Calculate the resultant force on Q₂ given the following charge configuration:
Q2Q1
Q3
0,05m
+8nC
FQ1 on Q2 = k𝑄1𝑄2
𝑟2 = 9 x 109 x (8×10−9)×(3×10−9)
(0,05)2
= 8,64 x 10-5 N, to the right
0,03m
-2nC
+3nC
Additional examples:4. Calculate the resultant force on Q2 given the following charge configuration:
FQ3 on Q2 = k𝑄1𝑄2
𝑟2 = 9 x 109 x (2×10−9)×(3×10−9)
(0,03)2
= 6 x 10-5 N, downwards
Q2Q1
Q3
0,05m
+8nC
0,03m
-2nC
+3nC
Additional examples:4. Calculate the resultant force on Q2 given the following charge configuration:
Q28,64 x 10-5 N
6 x 10-5 N
6 x 10-5 N
θ
(Fnet)2 = (8,64 x 10-5)2 + (6 x 10-5)2
Fnet = 1,05 x 10-4 N
tan θ = (6 x 10-5) ÷ (8,64 x 10-5)
θ = 34,78o
∴ Fnet = 1,05 x 10-4 N; 124,8o from the line connecting Q₂ with Q₃
Homework:
p. 224
nos. 20, 23 (change the mass to 0,1g), 28
Electric field
An electric field as a region in
space in which an electric
charge experiences a force.
Electric fields
The shape of an electric field is
determined by the charge that
causes it.
The direction of the electric field at a point is
the direction in which a positive test charge
would move if placed at that point.
Properties of electric field lines
• Start and end perpendicular to the surface
of a charged object.
• May never cross.
• Field lines close together – strong field;
field lines further apart – weaker field.
• Direction by convention.
• Surround charged object in 3 dimensions.
Uniform electric field (not syllabus)
Exists between two parallel plates which
are oppositely charged.
A small positive test charge experiences
the same electrostatic force at any point
in this electric field.
+
-
Point charges
There is no uniform electric field around a point charge.
The further away from the point charge, the weaker the
electrostatic force.
There is no charge or electric field inside a hollow
charged object.
+ -
Charges with irregular shapes (not syllabus)
The strongest field is around the sharpest point.
+
Electric field strength
q
F E =
E = Electric field strength in N.C-1
F = Force in N
q = Charge in C (test charge)
Electric field strength at a POINT
in an electric field is the force per
positive unit charge at that point.
Electric field strength
Electric field strength around
a point charge or charged object.
Field strength around a charge “Q” at
a distance “r” from charge “Q”.
2r
kQ E =
2r
1 α Een
Q α E
q
F E =
2r
kQ E =
q experiences
field strength
Q causes
field strength
Homework
p. 247
nos. 10.1, 12.4, 13, 18.1
, away from the 5μC