Electric Fields Flashcards
Direction of field lines
The direction in which a point positive test charge would move
Point charges have a radial field
Two plates have a uniform field
Electric field strength
Force per unit positive charge
E = F/Q this is the general equation
Where E is the electric field strength
NC^-1
Uniform fields
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
Force between point charges
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
Similarities between electric and gravitational fields
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
Differences between gravitational and electric fields
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
Deriving the formula for work done
E=F/Q E=dV/d
Fd=Q dV
W=Fd
W=Q dV
Absolute electric potential
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
Relationship between electric field strength and electric potential
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
Electric potential energy
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
Electric potential difference
For two points in an electric field the potential difference is the energy needed to move a unit charge between those points
Electric potential for a radial field
V = (1/4pi E0) X (Q/r)
Potential gradient
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
Conducting paper
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
Electrolytic tank
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
Capacitance in series and in parallel
In series 1/Ct = 1/C1 + 1/C2 etc
In parallel Ct = C1 + C2 …
When do force fields arise?
Interaction of:
Mass
Static charge
Moving charge
Capacitor equation - time to halve
= 0.69RC
Relative permittivity
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
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?
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
Describe how a beam of fast moving electrons is produced in the cathode ray tube of an oscilloscope
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
Two charges are 40mm apart, point P is 40mm from each of the charges (0.8nC)
What is the electric potential at P?
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
