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A-Level Physics - OCR A > M6, C2 Electric Fields > Flashcards

M6, C2 Electric Fields Flashcards

1
Q

What does the capacitance of a capacitor depend on

A

The permittivity of the material between the plates - how easy it is to generate an electric field in the material.
Also the dimensions of the capacitor

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2
Q

What does this equation mean?

C = (ε_0 X A) / d

A

When there’s a vacuum between the plates!!!

capacitance = permittivity in free space X area of the plates / separation of the plates

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3
Q

What equation can you use for the capacitance of a material which has a material in between them instead of a vacuum.

A

C = εA / d

ε = relative permittivity X permittivity of free space

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4
Q

define electric field strength

A

force per unit positive charge

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5
Q

what is an electric field

A

the region around an object where it can attract or repel other charges

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6
Q

what does this equation mean

E = F / Q

A 
E = electric field strength
F = force on the charged object
Q = charge of the object
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7
Q

what are the units of electric field strength

A

NC^-1

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8
Q

is electric field strength a scalar or vector quantity

A

vector quantity

points in the direction that a positive charge would move

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9
Q

Point charges have a ______ electric field.

A

radial

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10
Q

Draw radial electric fields for a positive point charge and one for a negative point charge.

A

Positive point charge:
arrows pointing outwards

Negative point charge:
arrows pointing inwards

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11
Q

what is a uniform electric field

how can you produce one

A

has the same electric field strength everywhere

by connecting 2 parallel plates to the opposite poles of a battery
the field lines point from the plate with the more positive potential to the plate with the less positive potential

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12
Q

What is Coulomb’s law in words?

A

The force between two point charges is:

  • directly proportional to the product of the charges
  • inversely proportional to the square of their distance apart
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13
Q

What equation is

F = Qq / 4πε_0r^2

A

Coulomb’s law

Force on the object = charges of the two objects / (4π X permittivity of free space X distance between them squared)

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14
Q

what is coulomb’s law and the equation for electric field strength an example of?

A

inverse square law

F α 1/r^2

F is inversely proportional to 1 / r^2

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15
Q

draw a diagram for a negatively charged and positively charged object showing r and F

A

page 134 of year 2 textbook

F pointing towards the other object on both of them because they’re attracting
r is between the centres of the objects

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16
Q

derive an equation for electric field strength in radial fields

A

E = F/Q and F = Qq / 4πε_0r^2

therefore

E = Q / 4πε_0r^2

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17
Q

draw a diagram for 2 positively charged objects showing r and F

A

page 134 of year 2 textbook

forces point away from each other as they’re repelling
r is between the centres of the objects

18
Q

draw a graph of electric field strength (y) against distance (x)

A

curve slope downwards

page 136 of year 2 textbook

19
Q

calculate the force acting between a proton and an electron

the distance between them is 3.3X10^-4 m

A 
Q = 1.6X10^-19 C
q = -1.6X10^-19 C
r = 3.3X10^-4 m
ε_0 = 8.85X10^-12

F = (-1.6X10^-19 X 1.6X10^-19) / (4π X 8.85X10^-12 X (3.3X10^-4)^2)

= -2.11 X10^-21 N

20
Q

The radius of a gold nucleus is about 7X10^-15m and it has 79 protons.
Estimate the electric field strength at the surface of the nucleus.

A

E = Q / 4πε_0r^2

= (79 X 1.6X10^-19) / (4π X 8.85X10^-12 X(7X10^-15)^2)

= 2.32X10^21 NC^-1

21
Q

what does this equation mean

E = V/d

A

only for parallel plates

electric field strength = potential difference between the plate / distance between the plates

22
Q

what can you determine about the units of electric field strength
(relating the 2 equations)

A

E = V/d and E = F/Q

so

1 Vm^-1 = 1NC^-1

23
Q

what are the main differences between electric and gravitational fields

A

Gravitational forces are always attractive. Electric fields an either be attractive or repulsive.

Objects can be shielded from electric fields, but not gravitational fields.

The size of an electric force depends on the medium between the charges. For gravitational forces, this makes no difference.

24
Q

compare gravitational / electric fields

a) property that creates the field
b) type of field produced
c) field strength
d) forces between the particles
e) type of field

A

a) mass / charge
b) attractive (direction of field towards object) / negative point charge (attractive field) positive point charge (repulsive field)
c) g=F/m / E=F/Q
d) F = -GMm/r^2 / F = Qq/4πε_0r^2
e) point mass, radial field / point charge, radial field

25
Q

An electron is fired from one plate to another which is parallel. The separation between the plates is 1.2cm and the pd across them is 3.6kV.
Calculate the acceleration of an electron between the plates.

A

E= V/d
= 3600 / 1.2X10^-2 = 3X10^5 Vm^-1

E = F/Q
F = 3X10^5 X 1.6X10^-19 = 4.8X10-14 N

F=ma
a = 4.8X10^-14 / 9.11X10^-31
= 5.3X10^16 ms^-2

26
Q

When a positively charged particle travels through an electric field the force acts on it in the ____ direction as the field lines.

A

same

27
Q

When a negatively charged particle travels through an electric field the force acts on it in the ____ direction as the field lines.

A

opposite

28
Q

explain what is happen to a charged particle as it travels through an electric field

A

There’s a force acting on the particle. The work done on this particle by this force increases its kinetic energy and causes it to accelerate at a constant rate in the direction of the force. If the particle’s velocity initially has a component at right angles to the field lines, this component will remain unchanged and the velocity in the direction will be uniform.
The combined effect of constant acceleration and constant velocity at right angles to one another is a curved path.

29
Q

define electric potential

A

the work done bringing a unit positive charge from a point infinitely far away to that point in the electric field

30
Q

At infinity, the electric potential of a unit charge will be ______.

A

zero

31
Q

derive the equation for electric potential energy

energy = Qq / 4πε_0r

A

multiply the electric potential by the value of charge
=Vq

= Qq / 4πε_0r

32
Q

when you move a unit charge in an electric field, what is being done to the force

A

changing the electric potential energy

so work is being done to the force

33
Q

for a radial electric field, plot a graph of force (y) against distance between charges (x)

A

inverse square law
so a curved slope downwards

pg. 140 of year 2 textbook

34
Q

what does this equation mean and give units

V = Q / 4πε_0r

A 
V = electric potential in volts
Q = point charge creating the electric field in C
r = distance from the point charge in m
35
Q

explain what electric potential is when the charge is positive and draw a graph for it with V on y axis and r on x axis

A

Electric potential is positive when the charge is positive and the force is repulsive.
On a V-r graph, it’s a curved slope downwards, V is initially positive and tends to zero as r increases towards infinity.

(pg. 139 of yr 2 textbook)

36
Q

explain what electric potential is when the charge is negative and draw a graph for it with V on y axis and r on x axis

A

Electric potential is negative when the charge is negative and the force is attractive.
On a V-r graph, electric potential is initially negative and tends to zero as r increases towards infinity.

(pg. 139 of yr 2 textbook)

37
Q

Where does the greatest value of electric potential occur

A

on the surface of the charge

electric potential decreases as the distance from the charge increases

38
Q

A positively charged particle is placed 0.035m from the centre of a sphere with a charge of +3.1µC and radius 0.02m.
If the particle is then repelled by 0.19m, what change in potential does it experience?

A

change in potential = final potential - initial potential

final potential = 3.1X10^-6 / 4πε_0(0.035 + 0.19)
=123887.0869

initial potential = 3.1X10^-6 / 4πε_0 X 0.035
=796416.9872

123887.0869 - 796416.9872 = -6.72529…X10^5 V

= -6.7X10^5 V

39
Q

A capacitor is made up of 2 parallel plates, each with an area of 12mm^2. The plates are separated by a 0.1mm thick piece of paper which has a relative permittivity of 2.7. Calculate the capacitance of this capacitor.

A

permittivity of paper = ε_0 X ε_r
= 8.85X10^-12 X 2.7 = 2.3895X10^-11 Fm^-1

C = εA /d
= (2.3895X10^-11 X 12X10^-6) / 0.1X10^-3
= 2.8674 X10^-12
= 2.9 X10^-12 F

40
Q

derive the equation for the capacitance of an isolated charged sphere

C = 4πε_0R

A

As it’s a charged sphere, you can assume all of its charge is at its centre and treat it like a point charge.

V = Q/C
V = Q / 4πε_0R
(R is the radius of the sphere)
cancel Q
Q/C = Q / 4πε_0R

1/C = 1/4πε_0R
rearrange for C

C = 4πε_0R

A-Level Physics - OCR A (20 decks)

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