Electric Fields

Question 1

Direction of field lines

Answer 1

The direction in which a point positive test charge would move

Point charges have a radial field

Two plates have a uniform field

Question 2

Electric field strength

Answer 2

Force per unit positive charge

E = F/Q this is the general equation

Where E is the electric field strength

NC^-1

Question 3

Uniform fields

Answer 3

E = V/d

Used to determine whether a particle is charged

A charged particle that enters an electric field at right angles to the field will feel a constant force parallel to the field lines - it’s direction will depend on the charge

Accelerates the particle so it follows a parabola

Question 4

Force between point charges

Answer 4

See equation sheet

Air can be treated as a vacuum when calculating force - force is dependent on the permittivity of he medium

For a charged sphere, charge may be considered to be at the centre

This is called Coulomb’s law

It’s an inverse square

Question 5

Similarities between electric and gravitational fields

Answer 5

Formulas - inverse square law

Field strengths - force per unit charge/mass

Field lines

Potential - gravitational potential and negative electric potential are the same

Equipotentials - for a uniform spherical mass will form a spherical surface

Work done

Question 6

Differences between gravitational and electric fields

Answer 6

Gravitational forces are always attractive whereas electric forces can be attractive or repulsive

At subatomic levels Gravity is negligible whereas the electrostatic force isn’t

Question 7

Deriving the formula for work done

Answer 7

E=F/Q E=dV/d

Fd=Q dV

W=Fd

W=Q dV

Question 8

Absolute electric potential

Answer 8

The electric potential energy that a unit positive charge (+1C) would have at that point

The work done per unit charge to move a small test charge from infinity to that point

Question 9

Relationship between electric field strength and electric potential

Answer 9

Electric field strength = - gradient of potential/distance graph

Change in potential = area under electric field strength against distance graph

E is a vector but V is a scalar

Question 10

Electric potential energy

Answer 10

The energy stored by a charge due to its position in an electric field

It equals the work done moving a charge from infinity to that position

Question 11

Electric potential difference

Answer 11

For two points in an electric field the potential difference is the energy needed to move a unit charge between those points

Question 12

Electric potential for a radial field

Answer 12

V = (1/4pi E0) X (Q/r)

Question 13

Potential gradient

Answer 13

The change of potential per unit change of distance in any given direction

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

The potential gradient is in the opposite direction to the lines of force in an electric field

Question 14

Conducting paper

Answer 14

Can be used to map out field lines of a 2D field

Each edge is oppositely charged

A voltmeter is used to measure the potential difference of different points on the paper

Points with the same voltage can be joined up to show equipotentials

Question 15

Electrolytic tank

Answer 15

Tank of positive and negative ions dissolved electrodes are used to create a field

A voltmeter can be used to find points of the same voltage

From this equipotentials and field lines can be mapped out

Question 16

Capacitance in series and in parallel

Answer 16

In series 1/Ct = 1/C1 + 1/C2 etc

In parallel Ct = C1 + C2 …

Question 17

When do force fields arise?

Answer 17

Interaction of:

Mass

Static charge

Moving charge

Question 18

Capacitor equation - time to halve

Answer 18

= 0.69RC

Question 19

Relative permittivity

Answer 19

The ratio of the capacitance of a capacitor with a given dielectric material and the capacitance of the same capacitor but without the dielectric

Er = C/C0

Question 20

If two charges are 30cm apart, with charges 3q and 5q at what point will the force on a charged particle placed in a field equal 0?

Answer 20

To solve this the field strengths need to be equated where one r is equal to x and the other 30 - x

The constants can be cancelled out leaving:

3q/x^2 = 5q/(30-x)^2

This can then be solved to find that x = 13cm

Question 21

Describe how a beam of fast moving electrons is produced in the cathode ray tube of an oscilloscope

Answer 21

Electrons are produced by thermionic emission

The cathode is heated by the electric current

The increase in temperature increases the kinetic energy of the electrons until they have enough energy to overcome the work function

The electrons are then accelerated by the electric field between the anode and the cathode

Question 22

Two charges are 40mm apart, point P is 40mm from each of the charges (0.8nC)

What is the electric potential at P?

Answer 22

The potential is equal to double the potential of each individual charge - no need to resolve as potential is a scalar

Calculating this results in an answer of 360V