Electro statistics
Measuring Electric
Fields
Learning
Objectives
• Define the electric field vector at a point in space as the
quotient of the electrostatic force vector acting on a
positive test charge at this point divided by the test
charge
• Electric field
Keywords
• Electric field lines
• Field vectors
• Coulomb’s law
Question
Question: Two spheres P and Q have an even
charge distribution of 10 x 10 -6 C and
20 x 10 6 C and their centers are separated by
12 cm. What is the size and direction the
electrostatic force acting on the particles?
Charges are both positive so the direction is
that the forces will be repelling each other
Example Question
Question: Two spheres P and Q have an even
charge distribution of 10 x 10 -6 C and
20 x 10 6 C and their centers are separated by
12 cm. What is the size and direction the
electrostatic force acting on the particles?
Answer:
qP = +10 x 10-6 C
qq = +20 x 10-6 C
K = 9 x 109 Nm2C−2
r = 12 cm / 100 = 0.12 m = 12 x 10-2 m
Fe = ?
Charges are both positive so the direction is
that the forces will be repelling each other
Electric Field
Charged particles exert a force on one another. An electric field is a measurable
effect generated by any charged object.
https://www.youtube.com/watch?v=S8TT1xXfJuA
To understand electric field, let’s understand the working of plasma Sphere!
Let’s understand Electric Field
Electric Field Strength
Electric Field strength (E) relates force (F) on a positive test charge with
the strength of the charge (𝑞 / ) by :
The magnitude of the electric field is measured in Newtons per
Coulomb, N/C.
We can use coulomb’s law to find F
Coulomb’s Law
Coulombs law allows us to work out the force of attraction or repulsion
between two point-charges.
𝑭=
•
•
•
•
𝐹 is the force (N)
𝑄 is the charge (C)
𝑟 is the separation between the charges (m)
𝜖0 is the permittivity of free space
𝑸𝟏 𝑸𝟐
𝟒𝝅𝝐𝟎 𝒓𝟐
F = k 𝑸𝟏 𝑸𝟐
𝒓𝟐
Field Vectors
Electric fields can be modelled using arrows to represent the field vectors at various
locations.
The length of the arrow represents the field strength and the direction of the arrow
represents the field direction.
Direction of the force on a negative and positive charge.
Remember positive charges are red, negative charges are blue
Field Vectors
Field Vectors
• The magnitude of the electric field vector is calculated as the force
per charge on any given test charge located within the electric field.
• Since electric field is a vector, the usual operations that apply to
vectors can be applied to electric field. That is, they can be added in
head-to-tail fashion to determine the resultant or net electric field
vector at each location. This is shown on the next slide
Review Questions
A positive test charge of 5.0 × 10−6 C is in an electric field that
exerts a force of 2.0 × 10−4 N on it. What is the magnitude of the
electric field at the location of the test charge?
A negative charge of 2.0 × 10−8 C experiences a force of 0.060 N
to the right in an electric field. What are the field’s magnitude
and direction at that location?
A positive charge of 3.0 × 10−7 C is in a field of 27 N/C directed
toward the south. What is the force acting on the charge?
Review Questions
A positive test charge of 5.0 × 10−6 C is in an electric field that
exerts a force of 2.0 × 10−4 N on it. What is the magnitude of the
electric field at the location of the test charge?
Using E = F/q = we get 2.0 × 10−4 / 5.0 × 10−6 = 40 N/C
A negative charge of 2.0 × 10−8 C experiences a force of 0.060 N
to the right in an electric field. What are the field’s magnitude
and direction at that location?
Using E = F/q we get 0.06 / 2.0 × 10−8 = 3 x 106 Left
A positive charge of 3.0 × 10−7 C is in a field of 27 N/C directed
toward the south. What is the force acting on the charge?
Using F = Q x E we get 3.0 × 10−7 x 27 = 8.1 x 10−6 N
Review Questions
What is the magnitude of the electric field at a position that is
1.2 m from a 4.2 x 10−6 C point charge? (Use 9 x 109 𝑁𝑚2 /𝐶 2 for
k)
What is the electric field at a position that is 1.6 m east of a point
charge of +7.2 × 10−6 C?
Review Questions
What is the magnitude of the electric field at a position that is
1.2 m from a 4.2 x 10−6 C point charge? (Use 9 x 109 𝑁𝑚2 /𝐶 2 for
k)
Using E = kq/𝑟 2 we get 9 x 109 x 4.2 x 10−6 / 1.22 = 26250 N/C
What is the electric field at a position that is 1.6 m east of a point
charge of +7.2 × 10−6 C?
Using E = kq/𝑟 2 we get 9 x 109 x 7.2 x 10−6 / 1.62 = 25313 N/C
East
Electric Field Lines
Field lines indicate the direction of the force due to the electric
field on a positive test charge.
Spacing between lines indicates the electric field’s strength.
Note:
1. Field lines never intersect each other.
2. The field lines are perpendicular to the charge
3. The magnitude of charge and number of field lines are
proportional to each other
4. The start point is at the positive charge and end is at the
negative charge
Electric Field Lines
Electric Field Lines
Electric Field Lines
Van De Graaff Generators
Van De Graaff Generators Working
Plenary LMS Quiz Electric Field