Electric Fields

Question 1

State the interactions between charged objects

Answer 1

Like charges repel

Opposite charges attract

Question 2

Describe electrical conductors and explain their properties

Answer 2

Electrical conductors conduct an electrical current

This is because they contain lots of free electrons which can move freely inside the conductor

Question 3

Describe how you would charge a metal

Answer 3

Making sure that the metal is isolated from the earth so that the charge given to it isn’t immediately neutralised by electrons transferring between the conductor and the Earth
The isolated conductor can be charged by contact with any charged objects

Question 4

Describe how a positively charged conductor loses its charge when in contact with the Earth

Answer 4

The ‘earthed’ electrons transfer from the Earth to the conductor to neutralise or discharge it

Question 5

Describe insulating materials and explain their properties

Answer 5

Electrical insulators do not conduct an electrical current

This is because they do not contain free electrons; all the electrons in the insulator are attached to individual atoms

Question 6

Describe the transfer of electrons when you rub an uncharged perspex rod with an uncharged cloth

Answer 6

Electrons are transferred from the rod to the cloth, so the cloth becomes negatively charged

Question 7

What makes a good electrical conductor?

Answer 7

materials (such as metals) which contain a lot of free electrons

Question 8

Why do insulating materials not conduct electricity?

Answer 8

Because they do not contain free electrons

Question 9

Describe and explain the shuttling ball experiment and give an equation for the current caused by the ball

Answer 9

• A conducting ball suspended by an insulated thread is between 2 vertical metal plates.
• If a high voltage is applied across the 2 plates, the ball is attracted to one plate, where on contact, will either gain or lose electrons, so will thus reverse its charge
• A current is therefore set up round the circuit since:
  I = Qf = Charge Q / Time for 1 cycle

Question 10

Define a field line

Answer 10

A field line in an electric field is a line which a small positive charge would move along if free to do so

Question 11

Describe the pattern of field lines for 2 oppositely charged points near each other

Answer 11

The field lines become concentrated at the points. A positive test charge released from an off-centre position would follow a curved path to the negative point charge

Question 12

Describe the pattern of field lines for 2 positively charged points near each other

Answer 12

The field lines each emerge from the positive charges, and where the field lines approach each other, they deflect each other away from each other

Question 13

Describe the pattern of field lines for a point object near an oppositely charged flat plate

Answer 13

The field lines are concentrated at the point object but they are at right angle to the plate where they meet. The field is strongest where the lines are most concentrated

Question 14

Describe the pattern of field lines for 2 oppositely charged flat plates

Answer 14

The field lines run parallel from one plate to the other, meeting the plates at right angles. The field is uniform between the plates because the field lines are parallel to each other

Question 15

Define electric field strength

Answer 15

The electric field strength, E, at a point in the field is the force per unit charge on a positive test charge placed at that point
E = F / Q

Question 16

Give the unit for electric field strength

Answer 16

NC⁻¹ (Newtons per Coulomb)
or
Vm⁻¹ (Volts per metre)

Question 17

Give the equation for the electric field strength between two charged plates

Answer 17

E = V / d

where V is the potential difference between the plates, and d is the distance between the plates

Question 18

Prove the equation E = V / d

Answer 18

Considering a small charge Q between the plates:
- the force F on charge Q is given by F = QE
- From moving the charge from the positive to the negative plate, the work done W by the field on Q is given by W = Fd = QEd
- Pd between the 2 plates is work done per unit charge so V = W / Q = QEd / Q = Ed
Therefore E = V / d

Question 19

Give the relationship between electric field strength and surface area

Answer 19

E ∝ Q / A
E = εₒQ / A
Where εₒ is the constant of proportionality

Question 20

Define electric potential

Answer 20

The electric potential at a certain position in any electric field is defined as the work done per unit positive charge on a ‘positive test charge’ (i.e. a small positively charged object) when it is moved from infinity to that position
Eᴘ = QV
where Eᴘ is the electric potential energy, and V is the electric potential at this position

Question 21

Where is the position of zero potential energy?

Answer 21

Infinity

Question 22

Give the units for electric potential

Answer 22

Volts (V) - equal to 1 JC⁻¹

Question 23

Define equipotentials

Answer 23

Equipotentials are lines of constant potential
e.g. a test charge moving along an equipotential has constant potential energy, so no work is done by the electric field on the test charge because the force is at right angles to the test charge

Question 24

At what angle do lines of force (field lines) cross the equipotential lines at?

Answer 24

90°

Question 25

Define potential gradient

Answer 25

The potential gradient at any position in an electrical field is the change of potential per unit charge of distance in a given direction

Question 26

If an electric field is uniform, such as between two oppositely charged parallel plates, how are the equipotentials between the plates spaced?

Answer 26

Equally

Question 27

If an electric field is non-uniform, how does the spacing of the equipotentials relate to the potential gradient?

Answer 27

The closer the equipotentials are, the greater the potential gradient is as right angles to the equipotentials

Question 28

Give the relationship between the electric field strength and the potential gradient

Answer 28

The electric field strength is equal to the negative of the potential gradient

Question 29

State Coulomb’s law

Answer 29

F = kQ₁Q₂ / r² = (1 / 4πε₀)(Q₁Q₂ / r²)

The force F, between 2 ‘point charges’, Q₁ and Q₂ over distance r, where k is the constant of proportionality

Question 30

Give the equation for the constant of proportionality and its units

Answer 30

k = 1 / 4πε₀

Units: Fm⁻¹

Question 31

Describe the similarities between Newton’s law of gravitation and Coulomb’s law of force between two point charges

Answer 31

They are both inverse square laws

Question 32

Give the equation for the surface charge density needed on the surface of a conducting sphere in air to produce an electric field strength of E at the surface

Answer 32

Q / A = ε₀E

where A = 4πr² = the surface area of a sphere of radius r

Question 33

Define a point charge

Answer 33

An expression for a charged object in a situation where distances under consideration are much greater than the size of the object

Question 34

Define a ‘test’ charge

Answer 34

A point charge in an electric field that does not alter the electric field in which it is placed.
Such alteration would happen if an object with sufficiently large charge is placed in an electric field and it causes a change in the distribution of charge that creates the field

Question 35

Give the equation for the force on a test charge q due to the electric field from a point charge Q at distance r away from each other

Answer 35

F = (1 / 4πε₀)(Qq / r²)

Question 36

Give the equation for the electric field strength of at distance r away from a point charge

Answer 36

E = F/q = Q / 4πε₀r²

Question 37

Describe why electric field strength is a vector quantity

Answer 37

If a test charge is in an electric field due to several point charges, each charge exerts a force on the test charge. The resultant force per unit charge (F/q) on the test charge gives the resultant electric field strength at the position of the test charge.

Question 38

Give the relationship between the resultant electric field strength for the forces of 2 point charges acting at right angles to each other on a test charge

Answer 38

E² = E₁² + E₂²
Because the forces are at right angles together, the resultant force can be worked out by Pythagoras’ formula to give
F² = F₁² + F₂²
Because E = F/q, the resultant electrical field strength is proportional to the resultant force

Question 39

Give the equation for the electrical potential at distance r from a point charge Q

Answer 39

V = Q / 4πε₀r

Question 40

Describe the relationship between the electric field strength and distance from a point charge

Answer 40

The electrical field strength (E) has an ‘inverse square’ relationship to the distance from the point charge (r)
E ∝ 1 / r²

Question 41

Describe the relationship between the electrical potential and distance from a point charge

Answer 41

The electrical potential (V) is inversly proportional to the distance from the point charge (r)
V ∝ 1 / r