16. Electrostatics

Physics 12 Honors/Regular – Notes – Electrostatics V2.0 (KWP) 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


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


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.


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.


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


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

16. Electrostatics

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