Electric Fields

Question 1

State Coulomb’s Law.

Answer 1

Coulomb’s Law states that the attractive or repulsive force acting between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Question 2

Define Electric Field strength.

Is it a scalar or vector?

Answer 2

Electric Field strength is the force experienced per unit charge at a specific point in an Electric Field.

VECTOR (because it’s an acceleration)

Question 3

Define Electric Potential.

Is it a scalar or vector?

Answer 3

Electric Potential is the work done per unit charge against a field (if repulsive) to move it from an infinite distance to that point in a field.

SCALAR (because its a representation of energy)

Note:

• The electric potential at an infinite distance for an attractive field is 0.
• If work is done by the field then the potential is negative.

Question 4

What do the arrows in the field diagrams represent?

Answer 4

Direction of arrows indicates the direction of travel for a positive test charge.
Density of arrows indicates Electric Field Strength at that point.

Question 5

What is an equipotential?

Answer 5

A line of points that all have the same potential to do work.

Question 6

Why do equipotential lines get further apart?

Answer 6

The equipotential surface gets further apart because as the distance from the charge increases the potential decreases.

Question 7

What does the graph of Electric Field Strength VS separation distance look like? (E VS r)

Answer 7

It looks like a 1/r^2 graph because:

𝐸 = 1/4𝜋𝜀0 * 𝑄/𝑟^2

• (1/4𝜋𝜀0) = k
• (Q is the charge of the field creator)

Remember E is a vector therefore will be above x axis if attractive and below if repulsive.

Question 8

How do you find the work done between points for a E VS r graph?

Answer 8

The area under will give the Work done per unit charge. Because W = F*s

Question 9

What does the graph of electric potential VS separation distance look like? (V VS r)

Answer 9

It looks like a 1/r graph because:

𝑉 = 1/4𝜋𝜀0 * 𝑄/r OR V = E*r

• (1/4𝜋𝜀0) = k
• (Q is the charge of the field creator)

Question 10

How do you find the field strength for a V VS r graph?

Answer 10

The gradient at any point will give us the Electric field strength. Because E = V/d
- d = r