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Physics 12 Honors/Regular – Notes – Electric Potential and Energy V1.0 (KWP) Page 1 of 6
For more information, refer to….
1. Giancoli - Physics: Principles with Applications (5th Edition): Chapter 17 – Electric Potential and Electric Energy; Capacitance
… as well as other resources.
Data Table Charge on proton/electron qp / e (±)1.60 × 10−19𝐶
Coulomb’s constant k 9.0 × 109 Nm2/C2 Permittivity of free space 𝜖0 = (1/𝑐2𝜇0) 8.85 × 10−12 C2/Nm2
Electron rest mass 𝑚𝑒 9.11 × 10−31 kg Proton rest mass 𝑚𝑝 1.6726 × 10−27 kg
Neutron rest mass 𝑚𝑛 1.6749 × 10−27 kg
In the diagram above,
Example
An electron close to the surface of the negative plate is
attracted by the positive plate.
Assuming that it starts at rest, calculate the electron’s
speed when it reaches the positive plate.
(e = )
+ -
+
B A
𝑊 = ∆ 𝐸𝑃
(EP increses)
∆ 𝑉
(+) work: Work done ON the particle
against the electric field
+
It is convenient to define a new term:
𝑉 = 𝐸𝑃
𝑞 , where 𝑉
is called the electric potential. It is
measured in J/C. (Or 𝐸𝑃 = 𝑞𝑉)
Charge of the particle (C)
𝑊 = ∆ 𝐸𝑃 = 𝑞 ∆𝑉
∆𝑉 is the potential difference between in
the positions A and B. ∆𝑉 is often called
the voltage across the device in circuits
which can be measured with a voltmeter. 𝑉
+ -
i f
e -
1000 V −1.60 × 10−19𝐶
Physics 12 Honors/Regular – Notes – Electric Potential and Energy V1.0 (KWP) Page 2 of 6
Solution: According to the conservation of energy,
or,
Aside
When v -> c, then we need to replace the “rest mass”, m0, of the particle with its relativistic
mass given by,
Refer to unit of Theory of Special Relativity (Giancoli - Physics: Principles with Applications
(5th Edition) - Chapter 26). This particle will never travel at a speed equal or greater than the
speed of light, c.
Electric Field across parallel plates
But
so
Substitute equation (2) into equation (1):
Uniform electric field,
Only for Parallel Plates
∆ 𝐸𝑃 = −∆ 𝐸𝐾 |∆ 𝐸𝑃| = |∆ 𝐸𝑃|
𝑞∆𝑉 = −(𝐸𝑘𝑓 − 𝐸𝑘𝑖) 0
𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔
(−)
(−1.60 × 10−19𝐶)(1000𝑉) = −1
2𝑚𝑣𝑓
2
The electric potential
energy of the
negatively charged
particle decreases as it
is moved from the
negative plate to the
positive plate.
−1.60 × 10−16 = −1
2(9.11 × 10−31𝑘𝑔)𝑣𝑓
2
𝑣𝑓 = 1.8 × 107 𝑚/𝑠
(-) Work done BY the due to the field.
𝑚 =𝑚0
√1 −𝑣2
𝑐2
- +
+
q
d
∆𝑉
𝑊 = 𝑞 ∆ 𝑉
Uniform Electric Field (Constant force to move
the inside across)
𝑊 = 𝑞 ∆ 𝑉
𝐹𝑑 = 𝑞 ∆ 𝑉
E⃑⃑ =F𝑒⃑⃑⃑⃑
𝑞
F𝑒⃑⃑⃑⃑ = E⃑⃑ 𝑞
(E⃑⃑ 𝑞) 𝑑 = 𝑞 ∆ 𝑉
(1)
(2)
E⃑⃑ = ∆ 𝑉
𝑑
Physics 12 Honors/Regular – Notes – Electric Potential and Energy V1.0 (KWP) Page 3 of 6
Thus we have shown that the uniform field between parallel, , can be found if we know
the potential difference and the separation of the plates.
(Note: We have also shown that E can also be measured in V/m.)
Example
Accelerating plates Deflecting plates
Work done and point charge sources
E⃑⃑
- +
1000𝑉
(a) Draw the electric field lines between the plates.
Solution: Already drawn on the diagram.
(b) Calculate the electric field, E between the plates.
Solution:
Constant force Constant acceleration Uniform motion Equations of motion
E⃑⃑ =|∆𝑉|
𝑑=
|1000𝑉|
1.5 × 10−2𝑚= 6.7 × 104 𝑉/𝑚(𝑙𝑒𝑓𝑡)
1.5 cm
+ -
+
-
Parabolic
Compare with projectile motion
e-
e-
In gravitational we learnt that potential energy is
given by the equation 𝐸𝑃 = −𝐺𝑀𝑚
𝑟. This has been
derived by finding the area under the graph.
Similarly the potential energy of a charged
particle, q, at a distance r, from a point charge
source Q, is given by
𝐸𝑃 =𝑘𝑄𝑞
𝑟
𝐹𝐺
𝑟 0
Physics 12 Honors/Regular – Notes – Electric Potential and Energy V1.0 (KWP) Page 4 of 6
Example 1
Calculate the change in potential energy (or work done) to move the electron from A to B.
Or
Note: The work done to move the electron from A to B is equal to the change in potential
energy.
Energy and point charges
𝐹
𝑟 0
𝑄 𝑞 r
𝐸𝑃 =𝑘𝑄𝑞
𝑟
𝐹𝑖𝑥𝑒𝑑 𝑃𝑜𝑖𝑛𝑡 𝑐ℎ𝑎𝑟𝑔𝑒 + 5𝜇𝐶
𝐴 𝐵
e-
2 𝑐𝑚
3 𝑐𝑚
e = −1.60 × 10−19𝐶
∆𝐸𝑃 = 𝐸𝑃𝐵− 𝐸𝑃𝐴
= (𝑘𝑄𝑞
𝑟)𝐵 − (
𝑘𝑄𝑞
𝑟)𝐴
= 𝑘𝑄𝑞 (1
𝑟𝐵−
1
𝑟𝐴)
= (9.0 × 109 Nm2
C2 )(+5.0 × 10−6C)(−1.60 × 10−19𝐶) (1
0.03𝑚−
1
0.02𝑚)
= 1.2 × 10−13 𝐽
∆𝐸𝑃 = 𝑞∆𝑉 = 𝑞(𝑘𝑄
𝑟𝐵−
𝑘𝑄
𝑟𝐴)
Electric potential of a point charge
+ e
- Q1
The work done to move the e- from a distance
ri and rf away from the fixed point charge Q1 is
𝑟𝑓
𝑟𝑖
W = ∫ 𝐹 𝑑𝑟𝑟𝑓
𝑟𝑖
𝑊
𝑟𝑓 𝑟𝑖
where F =𝑘𝑄𝑒
𝑟2
Charge of an electron −𝟏. 𝟔𝟎 × 𝟏𝟎−𝟏𝟗𝑪
This gives ∆W =𝑘𝑄𝑒
𝑟= ∆E𝑃, (if 𝑟𝑓 𝑖𝑠 ∞ 𝑎𝑛𝑑 𝑟𝑖 𝑖𝑠 𝑟)
𝑟𝑓
𝐸𝑃𝑓
− 𝐸𝑃𝑖=
𝑘𝑄𝑒
𝑟
Solution:
Physics 12 Honors/Regular – Notes – Electric Potential and Energy V1.0 (KWP) Page 5 of 6
Don’t forget that charge of an electron is negative! This negative sign would cancel out the
other negative sign in the equation.
In general, the electric potential energy of a charged particle q that is at a distant r away from
a point charge Q is given by the following equation:
It is again convenient to use electric potential, V, which is
… due to a point charge Q. (Not for parallel plates: )
Example 1
𝐸𝑃𝑖= −
𝑘𝑄𝑒
𝑟
Charge of an electron −𝟏. 𝟔𝟎 × 𝟏𝟎−𝟏𝟗𝑪 0 − 𝐸𝑃𝑖
=𝑘𝑄𝑒
𝑟
𝐴𝑠 𝑟 (𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒) → ∞, 𝐸𝑃𝑓 (𝑓𝑖𝑛𝑎𝑙 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦) → 0
𝐸𝑃 =𝑘𝑄𝑞
𝑟
Note: 𝐸𝑃∞= 0
𝑘 is the Coulomb’s constant
𝐸𝑃 (J)
0 𝑟 (m)
Same polarity (i.e.: [+Q, +q] or [-Q,-q])
Opposing polarity (i.e.: [+Q, -q] or [-Q, +q])
… compare with gravitational attraction
𝑉 =
𝐸𝑃
𝑞
𝑉 =
𝑘𝑄
𝑟
∆𝑉 = �⃑� 𝑑
𝑄1 = 1𝜇𝐶
𝑄2 = −2𝜇𝐶
+ −
𝐴 𝐵
5 𝑐𝑚
4 𝑐𝑚
6 𝑐𝑚
p+
𝑄1 𝑄2
Physics 12 Honors/Regular – Notes – Electric Potential and Energy V1.0 (KWP) Page 6 of 6
Calculate:
(a) The proton’s change in potential, , as it moves from A to B. (b) The work done to get proton from A to B. Note:
“-“ sign must be used in the equations and should not be omitted. Energy (and therefore) potential are scalar. Don’t worry about direction. (There are no
components)
Solution: (a) To solve this question, we need to solve for the electric potential of the proton at point A and point B, and then find change in electric potential by taking the difference.
(b)
∆𝑉
𝑉1𝐴 =𝑘𝑄1
𝑟1𝐴
𝑉2𝐴 =𝑘𝑄2
𝑟2𝐴
𝑉𝐴 = 𝑉1𝐴 + 𝑉2𝐴 =
𝑘𝑄1
𝑟1𝐴+
𝑘𝑄2
𝑟2𝐴
= (9.0 × 109Nm2
C2)(
1 × 10−6 C
9 × 10−2 m)(
−2 × 10−6 C
4 × 10−2 m) = −3.5 × 105 V
𝑉1𝐵 =𝑘𝑄1
𝑟1𝐵
𝑉2𝐵 =𝑘𝑄2
𝑟2𝐵
𝑉𝐵 = 𝑉1𝐵 + 𝑉2𝐵 =
𝑘𝑄1
𝑟1𝐵+
𝑘𝑄2
𝑟2𝐵
= (9.0 × 109
Nm2
C2)(
1 × 10−6 C
15 × 10−2 m)(
−2 × 10−6 C
10 × 10−2 m) = −1.2 × 105 V
∆𝑉 = 𝑉𝐵 − 𝑉𝐴 = (−1.2 × 105 V) − (−3.5 × 105 V) = 2.3 × 105 V
𝑊 = 𝑞∆𝑉
= 𝑞∆𝑉 = (𝟏. 𝟔𝟎 × 𝟏𝟎−𝟏𝟗𝑪)(2.3 × 105 V)
= 3.7 × 10−14𝐽
Note: Charge of proton
has same magnitude as
that of the electron.
An increase in 𝐸𝑃