# Electric Fields Flashcards Preview

## Physics > Electric Fields > Flashcards

Flashcards in Electric Fields Deck (22)
1
Q

Direction of field lines

A

The direction in which a point positive test charge would move

Point charges have a radial field

Two plates have a uniform field

2
Q

Electric field strength

A

Force per unit positive charge

E = F/Q this is the general equation

Where E is the electric field strength

NC^-1

3
Q

Uniform fields

A

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

4
Q

Force between point charges

A

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

5
Q

Similarities between electric and gravitational fields

A

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

6
Q

Differences between gravitational and electric fields

A

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

7
Q

Deriving the formula for work done

A

E=F/Q E=dV/d

Fd=Q dV

W=Fd

W=Q dV

8
Q

Absolute electric potential

A

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

9
Q

Relationship between electric field strength and electric potential

A

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

10
Q

Electric potential energy

A

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

11
Q

Electric potential difference

A

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

12
Q

Electric potential for a radial field

A

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

13
Q

A

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

14
Q

Conducting paper

A

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

15
Q

Electrolytic tank

A

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

16
Q

Capacitance in series and in parallel

A

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

In parallel Ct = C1 + C2 …

17
Q

When do force fields arise?

A

Interaction of:

Mass

Static charge

Moving charge

18
Q

Capacitor equation - time to halve

A

= 0.69RC

19
Q

Relative permittivity

A

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

20
Q

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?

A

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

21
Q

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

A

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

22
Q

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

What is the electric potential at P?

A

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