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Uniform Electric Field (Motion Parallel to the Electric Field)
Uniform Electric Field (Motion Parallel to the Electric Field)
In the previous section we looked qualitatively at uniform Electric fields. By combining this understanding with the equations of motion calculations of the previous unit, it is possible to describe mathematically how a charged particle will move in an Electric field.
In the previous section we looked qualitatively at uniform Electric fields. By combining this understanding with the equations of motion calculations of the previous unit, it is possible to describe mathematically how a charged particle will move in an Electric field.
By applying the formulae below, the motion of a charged particle can be found :-
By applying the formulae below, the motion of a charged particle can be found :-
E = V / d Fd = QV F = EQ F = ma v = u + at s = ut + 1/2at2 V2 = u2 +2as
E = V / d Fd = QV F = EQ F = ma v = u + at s = ut + 1/2at2 V2 = u2 +2as
Example 1 -
Example 1 -
The potential difference across a set of horizontal Parallel Plates is 2 kV. If the plates are separated by 25cm, calculate :-
The potential difference across a set of horizontal Parallel Plates is 2 kV. If the plates are separated by 25cm, calculate :-
1. The unbalanced force on an electron midway between the plates
1. The unbalanced force on an electron midway between the plates
2. The Acceleration of the Electron
2. The Acceleration of the Electron
3. The time to reach the top plate
3. The time to reach the top plate
Three stage calculation -
Three stage calculation -
a. Force up due to Electric field
a. Force up due to Electric field
b. Force down due to weight
b. Force down due to weight
c. Combine to find Unbalanced Force
c. Combine to find Unbalanced Force
.
.
E = V / d
E = V / d
= 2000 / 0.25
= 2000 / 0.25
= 8x103 Vm-1
= 8x103 Vm-1
Fup = EQ
Fup = EQ
= 8x103 x 1.6x10-19
= 8x103 x 1.6x10-19
= 1.3x10-15 N ( upwards )
= 1.3x10-15 N ( upwards )
Fdown = mg
Fdown = mg
= 9.11x10-31 x 9.81
= 9.11x10-31 x 9.81
= 8.9x10-30 N ( downwards )
= 8.9x10-30 N ( downwards )
Fun = Fup - Fdown
Fun = Fup - Fdown
≃ 1.3x10-15 N ( upwards )
≃ 1.3x10-15 N ( upwards )
F = ma
F = ma
a = F / m
a = F / m
= 1.3x10-15 / 9.11x10-31
= 1.3x10-15 / 9.11x10-31
= 1.4x1015 ms-2 ( upwards )
= 1.4x1015 ms-2 ( upwards )
Dist. to travel 12.5cm
Dist. to travel 12.5cm
0.125 = 0 + 1/2 x 1.4x1015 x t2
0.125 = 0 + 1/2 x 1.4x1015 x t2
t = 1.3x10-8 s
t = 1.3x10-8 s
Energy transformations associated with moving charges
Energy transformations associated with moving charges
When a charged particle is moved within an Electric field, Work is done on it. The energy associated with this Work Done is equal to the gain in the Kinetic Energy of the particle.
When a charged particle is moved within an Electric field, Work is done on it. The energy associated with this Work Done is equal to the gain in the Kinetic Energy of the particle.
Work done to move charge = F x d
Work done to move charge = F x d
Kinetic Energy gained = 1/2 mv2
Kinetic Energy gained = 1/2 mv2
Fd = 1/2mv2
Fd = 1/2mv2
v = (2QV / m)0.5
v = (2QV / m)0.5
Example 2 -
Example 2 -
An electron is accelerated from rest through a potential of 50 V. What is its final velocity?
An electron is accelerated from rest through a potential of 50 V. What is its final velocity?
v = ( ( 2 x 1.6x10-19 x 50) / 9.11x10-31 )0.5
v = ( ( 2 x 1.6x10-19 x 50) / 9.11x10-31 )0.5
v = 4.19 x106 ms-1
v = 4.19 x106 ms-1
Note - In the above calculation, the final Velocity is very fast but is still understood using classical mechanics. If the velocity was greater than ~ 0.1c then relativistic effects must be taken into account (see unit one).
Note - In the above calculation, the final Velocity is very fast but is still understood using classical mechanics. If the velocity was greater than ~ 0.1c then relativistic effects must be taken into account (see unit one).
Uniform Electric Field ( Motion Perpendicular to the Electric Field )
Uniform Electric Field ( Motion Perpendicular to the Electric Field )
In the above section, the charged particle was either stationary to start with, or was moving parallel to the electric field, causing the charged particle to be accelerated.
In the above section, the charged particle was either stationary to start with, or was moving parallel to the electric field, causing the charged particle to be accelerated.
If, however, the charged particle enters the Electric field at a right angle then the particle behaves very differently.
If, however, the charged particle enters the Electric field at a right angle then the particle behaves very differently.
When the particle enters perpendicular to the field, the Force it experiences due to the field is at right angles to the motion, causing the charged particle to follow a parabolic path.
When the particle enters perpendicular to the field, the Force it experiences due to the field is at right angles to the motion, causing the charged particle to follow a parabolic path.
Note - This motion is analogous to an object falling under the influence of Gravity, the motion horizontally and vertically are independent of each other.
Note - This motion is analogous to an object falling under the influence of Gravity, the motion horizontally and vertically are independent of each other.
In order to calculate the effect on a charged particle moving in an Electric field, it can be treated in a similar way to a object following Projectile Motion. The only additional step is to determine a value of vertical Acceleration.
In order to calculate the effect on a charged particle moving in an Electric field, it can be treated in a similar way to a object following Projectile Motion. The only additional step is to determine a value of vertical Acceleration.
Example 3 -
Example 3 -
In the above set up, an Electron enters an Electric field midway between two charged plates, travelling at 2.1x107 ms-1. The plates are separated by a distance of 50mm and are 80mm long. If there is a Potential Difference of 500 V across the plates, calculate :-
In the above set up, an Electron enters an Electric field midway between two charged plates, travelling at 2.1x107 ms-1. The plates are separated by a distance of 50mm and are 80mm long. If there is a Potential Difference of 500 V across the plates, calculate :-
1. The time taken for the Electron to pass through the plates
1. The time taken for the Electron to pass through the plates
2. The vertical deflection (S) as it leaves the plates
2. The vertical deflection (S) as it leaves the plates
The first question is a simple speed, distance, time question:-
The first question is a simple speed, distance, time question:-
t = d/v = 80x10-3 / 2.1x107 = 3.8x10-9 s
t = d/v = 80x10-3 / 2.1x107 = 3.8x10-9 s
To find the vertical deflection, the acceleration and therefore
To find the vertical deflection, the acceleration and therefore
Force vertically must be found, before applying an equation of motion to find displacement :-
Force vertically must be found, before applying an equation of motion to find displacement :-
F = EQ = V/d x Q = ( 500x1.6x10-19 ) / ( 50x10-3 ) = 1.6x10-15N
F = EQ = V/d x Q = ( 500x1.6x10-19 ) / ( 50x10-3 ) = 1.6x10-15N
a = F/m = 1.6x10-15 / 9.11x10-31 = 1.756x1015 ms-2
a = F/m = 1.6x10-15 / 9.11x10-31 = 1.756x1015 ms-2
s = ut + 1/2at2 = 0 + 0.5x( 1.756x1015 ) x ( 3.8x10-9 )2 = 1.27x10-2 m
s = ut + 1/2at2 = 0 + 0.5x( 1.756x1015 ) x ( 3.8x10-9 )2 = 1.27x10-2 m
Non-Uniform Electric Field ( Head-On approach )
Non-Uniform Electric Field ( Head-On approach )
In the above situations, the uniform Electric field allowed simple calculations based using the equations of motion. If, However, a non-uniform field is present, it becomes much easier to characterise the situation as an energy conservation problem.
In the above situations, the uniform Electric field allowed simple calculations based using the equations of motion. If, However, a non-uniform field is present, it becomes much easier to characterise the situation as an energy conservation problem.
In the above diagram of the Rutherford Scattering experiment, Alpha particles were fired towards a sheet of thin Gold leaf.
In the above diagram of the Rutherford Scattering experiment, Alpha particles were fired towards a sheet of thin Gold leaf.
In this case, we are only concerned with the particles whose path is straight towards the Gold nucleus, which are deflected through 180 degrees (The top Alpha Particle in the above diagram).
In this case, we are only concerned with the particles whose path is straight towards the Gold nucleus, which are deflected through 180 degrees (The top Alpha Particle in the above diagram).
As both the Alpha particle and the Gold nucleus have a positive charge, the closer they become to each other, the greater the repulsion experienced.
As both the Alpha particle and the Gold nucleus have a positive charge, the closer they become to each other, the greater the repulsion experienced.
As the Alpha particle approaches, it experiences a Force opposing its motion. This Force causes the Alpha particle to lose Kinetic Energy. This Ek is converted into Electrostatic Potential Energy, giving a point of closest approach when Ek = 0 and Ep= max :-
As the Alpha particle approaches, it experiences a Force opposing its motion. This Force causes the Alpha particle to lose Kinetic Energy. This Ek is converted into Electrostatic Potential Energy, giving a point of closest approach when Ek = 0 and Ep= max :-
By conservation of Energy, a value for closest approach can be obtained :-
By conservation of Energy, a value for closest approach can be obtained :-
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