17. Electric Potential and Energy

Physics 12 Honors/Regular – Notes – Electric Potential and Energy V1.0 (KWP) 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


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


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


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


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:


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


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


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

17. Electric Potential and Energy

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