D2 - Electric and Magnetic Fields Flashcards
what are the steps to set up an electric fields demo
- set up a high potential difference across 2 plates
- sprinkle dipolar grains (semolina) on castor oil (this allows the grains to move)
- observe the movement of the grains
what are the 2 types of electric fields
radial
uniform plates
what is the strength for a uniform plate
the field has a constant field strength everywhere, because the distance between the plates are always the same
which way do the lines go from in uniform plates
from positive to negative
what are the 3 things you need to draw to get the three marks for a uniform plate drawing
- the field line between the plates should be equally spaced
- there should be arrows on the field lines going from positive to negative with the labelled plates
- you should have more than 3 field lines drawn between the plates
how do you draw the field lines for a positive radial charge
- draw a positively charged particle in the centre
- draw about 12 lines coming out of the particle at equal intervals (it should look a bit like a spider diagram)
- add the arrows going out of the particle
where is the force strongest on a radial charge
it is strongest closer to the particle
how do you draw the field lines for a negative radial charge
- draw a negatively charged particle in the centre
- draw about 12 lines coming out of the particle at equal intervals (it should look a bit like a spider diagram)
- add the arrows going in of the particle
where is the force weakest on a radial charge
it is weakest further away from the particle
what is the definition of electric field strength
the force per unit charge on a small positive test charge
can a positive charge survive without a negative charge in an electric field? how about for a magnetic field
electric field - positive can live without negative charge
magnetic - need both positive and negative poles to create a magnetic field
(EXAM Q)
Electric fields are caused by both point charges and by parallel plates with a pd across them. Describe the difference between the electric field caused by a point charge and the electric field between the parallel plates. your answer should include a diagram of each type of field and a reference to electric field strength
An electric field caused by a point charge has a stronger electric field strength closer to the point charge, that becomes weaker further away from charge because it is more spread out. The electric field caused by parallel plates is uniform, as the distance between the plates is constant. Therefore the electric field is not stronger or weaker at any specific point
Diagram drawing:
Parallel plates - draw 4 lines between two thin rectangles (the plates). Draw a positive sign on one plate and a negative on the other. Draw arrows going from positive to negative on the field lines
Point charge - draw a positive charge with 8-10 lines coming out of it perpendicular to the surface of the charge. Draw arrows coming out of the charge on the field lines
Whats the difference between a field force and a contact force
contact - physical contact
field - an object only experiences this when in the charges field
what does the electric field for an electron look like
A negative charge with field lines going into in. With arrows on the lines going into the charge
which charge do we use when calculating electric field strength
the charge affected by the force
What does an electric field between 2 parallel plates depend on
- the potential difference between the plates (V)
- the distance between them (d)
what is the formula for electric field strength between parallel plates
E = V/d
what is the charge of an electron? proton?
-1.6x10^-19C
1.6x10^-19C
what is the formula for potential energy for an electron
Ep = VQ
A charged particle of 8x10^-19C is held stationary inside an electric field produced by two electric plates. The voltage between the plates is 270V and they are separated by a distance of 6x10^-3m.
a) what constant field strength exists between the plates?
b) What is the mass of the particle? (Hint: first draw a free-body diagram of the particle to determine its weight)
a) E = V/d
E = 270/ 6x10^-3
E = 45000NC^-1
b) ‘stationary’ means the Weight equals the electric force.
Electric force = EQ
mg = EQ
m = EQ/g
m = (45000)(8x10^-19)/9.81
m = 3.7x10^-15 kg
A proton travelling at 3.4 x 10^6 ms^-1 passes through an electric field. How fast will the proton be going after it emerges from the field?
Ep = VQ
Ep = 1.2 x 10^-15J
Ke = 1/2mv^2
Ke = 9.7 x 10^-15J
Ep + Ke = 1.09 x 10^-14J
v = square root (2VQ/m)
because Ep = VQ (total energy),
v = square root (2Ep/m)
v = square root (2)(1.09 x 10^-14)/1.673 x 10 ^-27
v = 3.6 x 10^6 ms^-1
An electron travelling at 4.6 x 10^7ms^-1 enters a constant electric field between two charged plates spread 0.0085m apart. The voltage between the plates is 650V and the plates are 0.036m long.
a) what is the electric force acting on the electron
b) how much time is taken for the electron to pass through the plates
c) how far will the electron fall in its path, while in between the two plates
a) V/d = 76470.588 = E
F = E x q = 1.22 x 10 ^-4N
b) time = distance/ speed
time = 0.036/ 4.6 x 10^7
time = 7.83 x 10^-10
c)
s ?
u 4.6 x 10^7
v X
a F/m = 1.22 x10^-14/ 9.1 x 10^-31 = 1.34 x 10^16
t 7.83 x 10^-10
s = ut +1/2at^2
s = 0.04m
what is coulombs law
The electrical force between two stationary point charges where the force is inversely proportional to the square of the separation
what is a point charge
a particle of zero size that carries an electric charge
what is coulombs law equation
F = k(q1q2)/r^2
the force on q1 is …… in direction to the force on q2
equal in magnitude and opposite in direction
The force of 2 on 1 is equal but opposite to the force of 1 on 2
what is electric potential energy + formula
the potential energy a charged particle has, due to the force it experiences from the field
Ep = kQq/r
how do we find the work done to separate two attracting charges?
you need to find the electric potential energy at the start point and at the end point and minus them
Find the work done to move a charge (q) from 3m to 0.5m under the influence of charge (Q)
charge q = 3 x 10^-6
charge Q = 20 x 10^-6
Ep 1 = k(20x10^-6)(3x10^-6)/3
Ep 2 = k(20x10^-6)(3x10^-6)/0.5
Ep1 - Ep2 = 0.899J = 0.9J
What is electric potential (V) + formula
the electrical potential energy per unit charge at a point in a field
V = Wd/q = kQ/r
what is the electric potential at a point 2m from a +3C charge
V = kQ/r
V = (8.99x10^9)(3)/2
V = 1.35x10^10JC^-1
what is the potential halfway between a +3C charge and a -3C charge separated by 1m
No potential
Find the electric potential at point B due to the other charges:
A = +30x10^-6
C = +70x10^-6
distance between A and B = 20cm
distance between B and C = 60cm
V1 = kq/r (A)
V1 = 1348500
V2 = kq/r (A)
V2 = 1048833.333
V1 + V2 = 2.4x10^6V
how much energy does it take to move along an equipotential line
no energy
what is the gradient of the graph of electric potential
electric field strength
