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Physics 12 Honors/Regular – Notes – Electrostatics V2.0 (KWP) Page 1 of 6
For more information, refer to….
1. Giancoli - Physics: Principles with Applications (5th Edition): Chapter 16 – Electric Charge
and Electric Field
… as well as other resources.
Data Table Charge on proton/electron qp / e (±)1.60 × 10−19𝐶
Coulomb’s constant k 9.0 × 109 Nm2/C2 Permittivity of free space 𝜖0 = (1/𝑐2𝜇0) 8.85 × 10−12 C2/Nm2
Electron rest mass 𝑚𝑒 9.11 × 10−31 kg Proton rest mass 𝑚𝑝 1.6726 × 10−27 kg
Neutron rest mass 𝑚𝑛 1.6749 × 10−27 kg
Forces between charged particles
The electrostatic force between two charged particles is dependent on (1) The amount of charge on each particle ( F ∝ 𝑄1, 𝑄2 )
(2) The distance between the charges (F ∝1
𝑟2)
Coulomb’s Law
𝐹𝑒 =𝑘|𝑄1||𝑄2|
𝑟2
where k is Coulomb’s constant = 9.0 × 10
9 Nm
2/C
2
Q1 and Q2 are charges on particles 1 and 2 respectively, measured in C (Coulombs) 1 e
- (electron) has a charge 1.6×10
-19 C
r is the distance between particles, measured in m. We can determine the direction force by remember that
Question
Calculate the electrostatic force which a) Q1 acts on Q2
+ -
Q1=+1μC
10−6
Q2=-2 μC
5cm
Physics 12 Honors/Regular – Notes – Electrostatics V2.0 (KWP) Page 2 of 6
b) Q2 acts on Q1 Solution: a) & b)
𝐹𝑒 =𝑘|𝑄1||𝑄2|
𝑟2
𝐹𝑒 =9.0 × 109 Nm2/C2|+1 × 10−6𝐶||−2 × 10−6𝐶|
(0.05𝑚)2= 7.2 𝑁
a) Force that Q1 acts on Q2
F21 = 7.2 N left
b) Force that Q2 acts on Q1
F12 = 7.2 N right
“Electrostatic attraction” (c) If the positively charged particle was fired and the negatively charged particle has a mass of 1.68 × 10
-5 kg, calculate the acceleration of the negatively charged particle.
𝐹𝑛𝑒𝑡 = 𝑚𝑎 => 𝑎 = 7.2𝑁
1.68 × 10−5𝑘𝑔= 4.29 × 105
𝑚
𝑠2(𝑙𝑒𝑓𝑡)
Question Determine where electron would have a net force of zero (i.e. equilibrium position). We could solve it through a system of equations.
Solution: According to the Coulomb’s Law,
𝐹𝑒 =𝑘|𝑄1||𝑄2|
𝑟2𝐹𝑒 =
𝑘|𝑄1||𝑄2|
𝑟2
In this case
Since the electron has a net force of zero, F1=F2
Moreover, we know that
𝑟1 + 𝑟2 = 0.04
Newton’s third law
+
-
Q1
Q2
+ +
4cm
e F1 F2
2μC 3μC e=-1.6×10
-19 C
𝐹1 =𝑘|𝑄1||𝑒|
𝑟12 𝐹2 =
𝑘|𝑄2||𝑒|
𝑟22
∴𝑘|𝑄1||𝑒|
𝑟12=
𝑘|𝑄2||𝑒|
𝑟22 =>
2μ
𝑟12=
3μ
𝑟22 => 2𝑟2
2 = 3𝑟12 => 𝑟1
2 =2𝑟2
2
3
𝑟1 = 𝑟2√2
3 => 𝑟1 =
√6𝑟23
①
②
Physics 12 Honors/Regular – Notes – Electrostatics V2.0 (KWP) Page 3 of 6
Substituting equation (1) into (2), we obtain,
Electric field lines
A few other electric field lines diagrams… Note their differences!
√6𝑟23
+ 𝑟2 = 0.04 => (3 + √6
3) 𝑟2 = 0.04
𝑟2 = 0.0220 𝑚 = 2.20 𝑐𝑚
𝑟1 = 0.04 − 0.0220 = 0.0180 𝑚 = 1.80 𝑐𝑚
+
Stronger field
strength
Weaker field
strength… field
lines further apart
+
+ +
Use test charge (always positive “+”) to
determine the direction of the field, E⃑⃑ . Electric Field is the force acting in the field.
+
+
E⃑⃑ = F⃑⃑
𝑞
Electric field is measured in N/C.
q is the charge
of a charged
particle in the
field.
Around a point charge,
the electric field E is
not constant.
-
+
+
Test charge
Test charge
- + Fixed Charge
+
+
+
+
+ + Fixed Charge
+
+
+
*Field lines NEVER intersect.
Physics 12 Honors/Regular – Notes – Electrostatics V2.0 (KWP) Page 4 of 6
Charged Parallel Plates
The electric field strength, E⃑⃑ , acting upon a particle with a charge, q is given by:
Defines E (General)
so
- +
- +
- +
- +
- +
q
Uniform field (E⃑⃑ is constant)
Constant acceleration (Constant F⃑ )
means we can use the kinematics
equations to solve problems in this
field.
+
+
+
+
Fixed (+) charges
(+)
Fixed (-) charges
(+)
Since field lines are parallel and
equally spaced, the electric field
is constant.
Be careful that you’re consistent
in how you note your E.
E⃑⃑ , E𝑃 , ℇ
+
+
+
-
-
-
Parallel Field Lines
e
+
…
Uniform field (E⃑⃑ is constant) …
Constant acceleration
(Constant F⃑ )
+ -
F𝑒⃑⃑⃑⃑ = 𝑞E⃑⃑
E⃑⃑ =F𝑒⃑⃑⃑⃑
𝑞
Compare with F𝑔⃑⃑⃑⃑ = 𝑚𝑔
where q is the charge of the particle
placed in the electric field.
Physics 12 Honors/Regular – Notes – Electrostatics V2.0 (KWP) Page 5 of 6
Electric field due to a point charge source
=>
Electric field strength, E has units N/C (or as we will soon see later on, V/m)
Example
1. Where would I place an electron in the charge set up below so that it would remain in
equilibrium? (Draw a free body diagram of the electron.) – Assume Q1 and Q2 are fixed charges
here.
Solution:
For equilibrium,
Quadratic Equation:
E⃑⃑ =
𝑘𝑄𝑞𝑟2
𝑞 E⃑⃑ =
𝑘𝑄
𝑟2
Vector
Compare with g =𝐺𝑀
𝑟2 …
+ -
(Fixed) Q1=1μC (Fixed) Q2=-2μC
0.05m x
F2 <- e- -> F1
∑F = 0
|𝐹2| = |𝐹1|
𝑘|𝑄2||𝑒|
𝑟22=
𝑘|𝑄1||𝑒|
𝑟12
vector
𝑘|𝑄2||𝑒|
𝑟22=
𝑘|𝑄1||𝑒|
𝑟12
+
𝑟1
e Q1
e
-> F1
F2 <- - Q2
𝑟2
(2 × 10−6𝐶)
(𝑥 + 5 × 10−2𝑚)2=
(1 × 10−6𝐶)
𝑥2
(2 × 10−6)𝑥2 = (1 × 10−6)(𝑥 + 5 × 10−2)2
2𝑥2 = 𝑥2 + (1 × 10−1)𝑥 + 25 × 10−4
−𝑥2 + (1 × 10−1)𝑥 + 25 × 10−4 = 0
𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐
2𝑎
𝑥 =−(1 × 10−1) ± √(1 × 10−1)2 − 4(−1)(25 × 10−4)
2 × (−1)
𝑥 = +0.1207 m or 𝑥 = −0.0207 m
Physics 12 Honors/Regular – Notes – Electrostatics V2.0 (KWP) Page 6 of 6
In this case the negative value would be rejected. As a result, the electron should be placed
0.1207m away from the positive charge, and 0.1707 m away from the negative charge in
order to remain in equilibrium.
Question
(a) Solution:
(b) Solution:
+ -
e
An electron is placed within a pair of opposing charged
plates. Given that the uniform electric field between the
plates is 50 N/C, calculate
(a) The acceleration of the electron (Magnitude and
direction)
(b) The time it takes for the electron to move from one
plate to the other if the plates are 1.0 cm apart and the
electron starts from rest.
E⃑⃑ =F𝑒⃑⃑⃑⃑
𝑞
But since F⃑ = 𝑚a⃑ ,
E⃑⃑ =𝑚a⃑
𝑞
a⃑ =E⃑⃑ 𝑞
𝑚=
(50𝑁𝐶) ( 1.6 × 10−19𝐶)
(9.11 × 10−31𝑘𝑔)= 8.8 × 10−12𝑚/𝑠2, 𝑙𝑒𝑓𝑡
𝑑 = 𝑣𝑖𝑡 +1
2𝑎𝑡2
𝑡 = √2𝑑
𝑎= √
2 × 1.0 × 10−2𝑚
8.8 × 10−12𝑚/𝑠2= 4.8 × 10−8𝑠
0