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Electrostatics and Energy
Positive Work: Work was done ON the positive particle to move it against the
electric field.
It is convenient to define a new term,
π =πΈP
πor πΈP = ππ where π is measured in J/C.
π is called the electric potential, and π is the charge of the particle in C.
In the diagram to the left,
π = βπΈP = πβπ.
βπ is the potential difference between the two positions, A and B. It is often called the voltage across the
device in circuits, which can be measured with a voltmeter.
Example 1: An electron close to the surface of the negative plate is attracted by the
positive plate. Assuming that it starts at rest, calculate the electronβs speed when it
reaches the positive plate.
Due to the conservation of mechanical energy,
βπΈP = ββπΈK or |βπΈP| = |βπΈK|
β πβπ = β(πΈKf β πΈKi) = βπΈKf = β1
2ππ£π
2
β π£π = ββ2πβπ
π= ββ
2(β1.6 Γ 10β19 C)(1000 V)
9.11 Γ 10β31 kg= 1.9 Γ 107 m/s
Negative Work: Work was done BY the electron as it moves along the electric field.
Electric Field Across Parallel Plates
For a uniform electric field, constant force is needed to move the particle across.
π = πβπ = πΉπ where οΏ½βοΏ½ =πΉ
πor πΉ = οΏ½βοΏ½ π.
β΄ πβπ = (οΏ½βοΏ½ π)π β βπ = οΏ½βοΏ½ π
β΄ οΏ½βοΏ½ =βπ
π(only for parallel plates)
Thus, we have shown that the uniform electric field between parallel plates, οΏ½βοΏ½ ,
can be found if we know the potential difference and the separation of the plates.
(Note: we have also shown that οΏ½βοΏ½ can also be measured in V/m.)
Aside: When the velocity of a particle approaches the speed of light (π£ β π), we need to replace the βrest
massβ π0 of the particle with its relativistic mass, given by
π =π0
β1 βπ£2
π2
(Refer to the unit on Special Relativity, Ch. 26)
The particle will never travel at a speed equal to or greater than the speed of light, π.
Example 2: Calculate the electric field between the plates.
οΏ½βοΏ½ =βπ
π=
1000 V
0.015 m= 6.7 Γ 104 V/m left (toward the negative plate)
Accelerating Plates Deflecting Plates
(Compare with projectile motion)
Energy, Work, and Point Charges
We have previously learnt that gravitational potential energy is given by
πΈP = βπΊππ
π.
This was derived by finding the area under a graph of gravitational force, πΉG, vs. distance from Earthβs
centre, π. Now, we will use a similar process to determine the formula for electric potential energy.
Consider the diagrams below. The work done to move the electron from πi to πf, away from the fixed point
charge π, is
π = β« πΉππππf
πi
where πΉπ =πππ
π2
β π = β«πππ
π2ππ
πf
πi
= [βπππ
π]πi
πf
= βπππ (1
πfβ
1
πi)
1.5 cm
+ β
1000 V
πΉπ
π
π
πi πf
Since work is also equal to the change in potential energy,
βπΈP = πΈPf β πΈPi =βπππ
πfβ
βπππ
πi
As πf β β, πΈPf β 0 (πΈP decreases as we move farther away from π), so
0 β πΈPi =βπππ
ββ
βπππ
πi= 0 β
βπππ
πiβ πΈPi =
βπππ
πi
This electric potential energy will ultimately be positive: π, π, and πi are all positive, while π is negative
and will cancel out the negative sign in front to give a positive value overall.
In general, the electric potential energy of a charged
particle π at a distance π away from a point charge
source π is given by
πΈP =πππ
π
If π and π have the same charge/polarity (ie. both positive or both
negative), they will repel each other, and πΈP > 0.
If π and π have different charges/polarities (ie. one positive, one
negative), they will attract each other, and πΈP < 0.
As we get farther and farther away from the point charge π (ie. π
tends to infinity), the electric potential energy approaches zero.
That is,
πΈPβ = 0.
Again, notice the similarities between gravitational and electric
potential energy.
From our equation for electric potential energy, we can also write
πΈP =πππ
π= ππ β΄ π =
ππ
π
to represent the electric potential due to a point charge π.
Note: Energy and potential difference are scalar. Do not worry
about direction here (there are no components either).
Example 3: Calculate the work done to move the
electron from A to B.
Note: Negative signs must be used in the equation
and NOT OMITTED.
βπΈP = πβπ = π (ππ
πBβ
ππ
πA)
πΈP (J)
π (m)
Same
Polarity
Opposing
Polarities
0
or βπΈP = πΈPB β πΈPA = (πππ
πB) β (
πππ
πA) = πππ [
1
πBβ
1
πA]
= (9 Γ 109 Nm2/C2)(+5 Γ 10β6 C)(β1.6 Γ 10β19 C) [1
0.03 mβ
1
0.02 m] = 1.2 Γ 10β13 J
Example 4: Calculateβ¦
a) the protonβs change in potential as it moves from A to B.
πA = π1A + π2A =ππ1
π1A
+ππ2
π2A
=(9 Γ 109 Nm2/C2)(+1 Γ 10β6 C)
0.09 m+
(9 Γ 109 Nm2/C2)(β2 Γ 10β6 C)
0.04 m= β3.5 Γ 105 V
πB = π1B + π2B =ππ1
π1B
+ππ2
π2B
=(9 Γ 109 Nm2/C2)(+1 Γ 10β6 C)
0.15 m+
(9 Γ 109 Nm2/C2)(β2 Γ 10β6 C)
0.10 m= β1.2 Γ 105 V
βπ = πB β πA = (β1.2 Γ 105 V) β (β3.5 Γ 105 V) = 2.3 Γ 105 V
b) the work done to get the proton from A to B. (π = +1.6 Γ 10β19 C)
π = βπΈP = πβπ = (+1.6 Γ 10β19 C)(2.3 Γ 105 V) = 3.7 Γ 10β14 J
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