Topic 7a - Electric Fields Model Answers ( Unfinished)

Question 1

Explain how an insulating object can become charged

Answer 1

Electrons can be transferred from one object to another
Via friction
The object gaining electrons becomes negatively charged

Question 2

Describe what is meant by an electric field

Answer 2

A regions where a charged particle experiences a force

Question 3

Define electric field strength

Answer 3

The force per unit positive charge

Question 4

State the equation for electric field strength

Answer 4

E =F/q
Where E is the electric field strength
F is the electrostatic force
Q is the charge of the charge placed within the field

Question 5

State whether the following are vectors or scalars
Coulomb force
Electric field strength

Answer 5

Vector
Vector

Question 6

Describe the directions of electric field lines

Answer 6

Electric field lines point in the direction that a positive point charge would experience a force

Question 7

Draw the electric field around a positive point charge

Answer 7

Radial field
Electric field strength decreases with distance from charge (field lines become further apart)
Field lines meet charge at right angles to its surface
Arrows point in the direction that a positive point charge would experience a force (away from +)
See physical photos for diagram

Question 8

Draw the electric field around a negative point charge

Answer 8

Radial field
Electric field strength decreases with distance form charge (field Ines become further apart)
Field lines meet charge at right angles to its surface
Arrows point in the direction that a positive point charge would experience a force (towards -)

Question 9

Show how the electric field at vector A has been constructed

Answer 9

1. Draw the electric field vectors due to positive and negative charges
   Note these vectors are the same length because point A is the same distance from each charge
2. Draw the electric field vectors due to the positive and negative charges Note these

Question 10

Show how the electric field at X has been constructed

Answer 10

Identify the filed vectors due to charge 1 and 2
Note that E1 is shorter in length than E2 as the point we are considering is further from charge 1 than charge 2
Then vector diagram (see photos for answers)

Question 11

Draw the electric field lines between a positive point charge and a negative plate

Answer 11

See photos for answer

Question 12

Define uniform electric field

Answer 12

A field in which the electric field strength is constant everywhere

Question 13

Draw the electric field between two infinite parallel plates

Answer 13

Arrows from + to - (as this is the direction a positive charge would experience a force if placed in the field)
Field lines equidistant (as the electric field strength is represented by the closeness of field lines, and this is constant)
Field lines meet plates perpendicular to the plate
See ph9tos for answers

Question 14

State the equation for the electric field strength in a uniform field

Answer 14

E =V/d
Where E is the electric field strength in
V is the potential differene across the plates
D is the separation between the two parallel plates

Question 15

State Coulomb’s law

Answer 15

· F = kQq/r2
· Where F is the electrostatic force
· k is Coulomb’s constant = 8.99x109 Nm2C-2
· and r is the separation between the centres of the charges

Question 16

Derive the base units for k

Answer 16

k = Fr2/Qq
[k] = [F] [r]2/[Q]2 = Nm2C-2
Base units are: m, s, A, kg
So as N = kgms-2 and C = As
[k] = kgms-2m2A-2s-2
= kgm3s-4A-2

Question 17

23. Describe the relationship between force and separation between charges

Answer 17

· The force is inversely proportional to the square of the separation (inverse square law)

Question 18

See photos for question 24 and answer

Question 19

25. Derive the expression for the electric field strength in a radial field

Answer 19

F = kQq/r2
E = F/q (from the definition of field strength)
E = kQq/r2q
E = kQ/r2
Where k is the Coulomb constant, Q is the charge creating the field, r is the distance from the centre of the charge Q

Question 20

26. Prove that the relationship between electric field strength and separation as shown in this graph obeys the inverse square law ( see photos for graph

Answer 20

· If E and r follow the inverse square law, then Er2 should be a constant as E = constant/r2
· E1r12 = 11 x1011 x (3.6 x10-11)2 = 1.43 x10-9 NC-1m2
· E2r22 = 4 x1011 x (6.0 x10-11)2 = 1.44 x10-9 NC-1m2
· E3r32 = 2 x1011 x (8.4 x10-11)2 = 1.41 x10-9 NC-1m2
· Er2 is a constant to 2sf hence E and r follow the inverse square law

Question 21

27. Determine the resultant electric field strength at position X ( see photos for X

Answer 21

Let the 60 μC charge be Q1 and the -80 μC charge be Q2
E1 = kQ1/r12 = 8.99x109 x 60 x10-6 /(12x10-3)2 =3.75 x109 NC-1
E2 = kQ2/r22 = 8.99x109 x 80 x10-6 /(9x10-3)2 = 8.88x109 NC-1
Magnitude of resultant – use Pythagoras:
Etotal =√(3.75 x109 )2 + (8.88x109 )2 = 9.64 x109 NC-1
Direction – use trigonometry

θ = tan-1 (8.88x109 /3.75 x109)
= 67.10 below the horizontal
(See diagram in photos

Question 22

28. Define electric potential

Answer 22

· The work done per unit positive charge in bringing a charge from infinity to a point within the field

Question 23

29. Explain why the electric potential is positive for positive charges and negative for negative charges

Answer 23

· In bringing a positive test charge from infinity to a point within the field around a positive charge, work has to be done against the field
· This is because the test charge would experience a repulsive electrostatic force in the opposite direction to its displacement
· So the test charge would gain electrical potential energy as it moves closer to the field creating positive charge (transferring it from kinetic energy)
· Whereas when a positive test charge is brought closer to a negative field creating charge, the test charge would experience an attractive electrostatic force towards the negative field creator
· Therefore work would be done by the field
· The charge’s electrical potential energy would decrease (it would be transferred into kinetic energy) as it got closer to the negative charge

Question 24

30. Sketch a graph of potential against separation from a) a positive charge b) a negative charge

Answer 24

(See photos

Question 25

31. Describe the relationship between electric potential V and separation r

Answer 25

· Potential and separation are inversely proportional

Question 26

See photos for question

Answer 26

200V/0.05 = 400NC-1 Potential difference / separation

Question 27

33. State the equation showing the relationship between electric potential and field strength

Answer 27

· E = -ΔV/Δr · E is the electric field strength · ΔV is the change in potential for a change in distance Δr

Question 28

34. David Blaine wore a metal suit and was placed in a room with a charged sphere that caused ionisation at 1m. A scientist claims that David Blaine is completely safe inside the metal cage. Comment on this suggestion. (4 marks)

Answer 28

There is no potential difference across the cage or his body -Electric field strength inside the cage is zero (it is a Faraday cage) - This is because there is no potential gradient -Current conducts only on the surface of the cage – not inside

Question 29

See question in photos

Answer 29

E = ∆V/∆r = 200/0.05 = 4000 Vm-1

Question 30

36. Explain how to determine the electric field strength from a graph of potential against separation

Answer 30

· Electric field strength is the negative gradient of a potential, distance graph · If the potential, distance graph is a curve, then the tangent at a point should be drawn and the gradient of this should be found