Electrostatics

Forces Between Charged Particles

The electrostatic force between two charged particles is dependent on

1) the amount of charge on each particle:

�⃑�e ∝ 𝑄1, 𝑄2

2) the distance between the charges:

�⃑�e ∝1

𝑟2

Coulomb’s Law

�⃑�e =𝑘𝑄1𝑄2

𝑟2where

�⃑�e is the electrostatic force (measured in Newtons, N)

𝑘 is Coulomb's constant = 9.0 × 109 Nm2/C2

𝑄1 and 𝑄2 are charges on particles 1 and 2, respectively* (measured in Coulombs, C)

𝑟 is the distance between particles 1 and 2 (measured in metres, m)

*An electron has a charge of −1.6 × 10−19 C.

We can determine the direction of the force by remembering that

1) LIKE charges REPEL 2) UNLIKE charges ATTRACT

Note: �⃑�e =1

4𝜋𝜀0∙

|𝑄1||𝑄2|

𝑟2is used in some higher level textbooks,

where ε0 is the permittivity of free space = 8.85 × 10−12 C2/Nm2

Question 1: Calculate the electrostatic force which

a) 𝑄1 acts on 𝑄2

b) 𝑄2 acts on 𝑄1

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.

Solution

First, let us calculate the magnitude of the

electrostatic force acting between 𝑄1 and 𝑄2:

|�⃑�e| =𝑘|𝑄1||𝑄2|

𝑟2=

(9.0 × 109 Nm2/C2)|1 × 10−6 C||2 × 10−6 C|

(0.05 m)2= 7.2 N

a) �⃑�21 = 7.2 N left b) �⃑�12 = 7.2 N rightNote the application of Newton’s Third Law: 𝐹21 and 𝐹12 are equal in magnitude but opposite in direction.

Since the particles are moving toward each other, this is a case of “electrostatic attraction.”

c) By Newton’s Second Law,

�⃑� = 𝑚�⃑� ⇒ �⃑� =�⃑�

𝑚=

−7.2 N

1.68 × 10−5 kg= −4.3 × 105 m/s2 (ie. 4.3 × 105 m/s2 left)

Question 2: Determine where the electron would experience a net force of zero (ie. equilibrium position).

∑ �⃑� = �⃑�1 + �⃑�2 = 0 ⇒ |�⃑�1| = |�⃑�2| (ie. �⃑�1 and �⃑�2 have the same magnitude)

Let 𝑟 be the distance of the electron from the 2 𝜇C-particle and 4 − 𝑟 the distance from the 3 𝜇C-particle.

(9.0 × 109 Nm2/C2)(2 × 10−6 C)(1.6 × 10−19 C)

𝑟2=

(9.0 × 109 Nm2/C2)(3 × 10−6 C)(1.6 × 10−19 C)

(4 − 𝑟)2

(2 × 10−6 C)

𝑟2=

(3 × 10−6 C)

(4 − 𝑟)2⇒ (2 × 10−6 C)(4 − 𝑟)2 = (3 × 10−6 C)𝑟2

Let us drop the units for now and multiply everything by 106, just to make our computations easier (this

does not change the value of the equation).

(2)(4 − 𝑟)2 = (2)(16 − 8𝑟 + 𝑟2) = 32 − 16𝑟 + 2𝑟2 = 3𝑟2

𝑟2 + 16𝑟 − 32 = 0 ⇒ this is not factorable, so we use the Quadratic Formula

𝑟 =−16 ± √162 − 4(1)(−32)

2(1)= 1.8 cm and − 17.8 cm

We reject the negative solution because it does not make sense in the real world. Thus, the electron

experiences a net force of zero 1.8 cm right of the 2 𝜇C-particle, or 2.2 cm left of the 3 𝜇C-particle.

Electric Field Lines

Every charge has an electric field around it which exerts forces on the surrounding charges.

We use a small positive test charge (𝑞0 in the pictures on the next page) to determine the direction of the

field. A test charge is so small that it does not significantly alter the electric field being measured.

The electric field strength, �⃑⃑�, acting upon a particle placed in the electric field is given by

�⃑⃑� =�⃑�

𝑞where �⃑⃑� is measured in N/C, and 𝑞 is the charge of the particle.

The further apart the field lines, the weaker the field strength. As a result, we would expect the field strength

to increase as we move closer to the charge.

On a side note, compare the electric field strength with the gravitational field strength for an object of mass

𝑚 inside a gravitational field:

�⃑�G =𝐺𝑀𝑚

𝑟2= 𝑚�⃑� ⇒ �⃑� =

𝐺𝑀

𝑟2where �⃑� is measured in N/kg.

Note how each is defined: �⃑⃑� is the force per unit charge, while �⃑� is the force per unit mass.

Electric Field due to a Single Point Charge

The electric field strength around a single, specific point charge 𝑄 is given by

�⃑⃑� =�⃑�𝑒

𝑞=

𝑘𝑄𝑞𝑟2

𝑞=

𝑘𝑄

𝑟2where 𝑟 is the distance from the point charge to where �⃑⃑� is measured.

Fixed

Charge

Fixed

Charge

Electric Field around Multiple Fixed Charges

Field lines never intersect.

In a uniform electric field, the field lines are parallel and equally spaced. The electric field strength, the

electrostatic force, and the acceleration are all constant, which means that we can use the equations of

motion to solve problems in this field.

Charged Parallel Plates

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.

At equilibrium, ∑ �⃑� = 0 ⇒ |�⃑�2| = |�⃑�1|

|𝑘𝑄2𝑒

𝑟22 | = |

𝑘𝑄1𝑒

𝑟12 | ⇒ |

𝑄2

(𝑟1 + 0.05 m)2| = |𝑄1

𝑟12| ⇒ |

−2 × 10−6 C

(𝑟1 + 0.05 m)2| = |

1 × 10−6 C

𝑟12

|

Drop the units and cross-multiply: (2 × 10−6)𝑟12 = (1 × 10−6)(𝑟1 + 0.05)2

Simplify: 2𝑟12 = (𝑟1 + 0.05)2 = 𝑟1

2 + 0.1𝑟1 + 0.0025 ⇒ 𝑟12 − 0.1𝑟1 − 0.0025 = 0

Use the quadratic formula: 𝑟1 =−(−0.1) ± √(−0.1)2 − 4(1)(−0.0025)

2(1)= 0.12, −0.021

We reject the negative solution because it does not make sense in real life applications. Therefore, you

should place the electron about 12 cm from the positive particle and 17 cm from the negative particle.

Example 2: 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).

�⃑⃑� =�⃑�

𝑞=

𝑚�⃑�

𝑞⇒ �⃑� =

�⃑⃑�𝑞

𝑚=

(50 N/C)(1.6 × 10−19 C)

9.11 × 10−31 kg= 8.8 × 1012 m/s2 toward the positive plate

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.

𝑑 = �⃑�i𝑡 +1

2�⃑�𝑡2 =

1

2�⃑�𝑡2 ⇒ 𝑡 = √2𝑑

�⃑�= √

2(0.010 cm)

8.8 × 1012 m/s2= 4.8 × 10−8 s