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,

𝑉 =𝐸P

𝑞or 𝐸P = 𝑉𝑞 where 𝑉 is measured in J/C.

𝑉 is called the electric potential, and 𝑞 is the charge of the particle in C.

In the diagram to the left,

𝑊 = ∆𝐸P = 𝑞∆𝑉.

∆𝑉 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 electron’s speed when it

reaches the positive plate.

Due to the conservation of mechanical energy,

∆𝐸P = −∆𝐸K or |∆𝐸P| = |∆𝐸K|

⇒ 𝑞∆𝑉 = −(𝐸Kf − 𝐸Ki) = −𝐸Kf = −1

2𝑚𝑣𝑓

2

⇒ 𝑣𝑓 = √−2𝑞∆𝑉

𝑚= √−

2(−1.6 × 10−19 C)(1000 V)

9.11 × 10−31 kg= 1.9 × 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.

𝑊 = 𝑞∆𝑉 = 𝐹𝑑 where �⃑� =𝐹

𝑞or 𝐹 = �⃑� 𝑞.

∴ 𝑞∆𝑉 = (�⃑� 𝑞)𝑑 ⇒ ∆𝑉 = �⃑� 𝑑

∴ �⃑� =∆𝑉

𝑑(only for parallel plates)

Thus, we have shown that the uniform electric field between parallel plates, �⃑� ,

can be found if we know the potential difference and the separation of the plates.

(Note: we have also shown that �⃑� can also be measured in V/m.)

Aside: When the velocity of a particle approaches the speed of light (𝑣 → 𝑐), we need to replace the “rest

mass” 𝑚0 of the particle with its relativistic mass, given by

𝑚 =𝑚0

√1 −𝑣2

𝑐2

(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, 𝑐.

Example 2: Calculate the electric field between the plates.

�⃑� =∆𝑉

𝑑=

1000 V

0.015 m= 6.7 × 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

𝐸P = −𝐺𝑀𝑚

𝑟.

This was derived by finding the area under a graph of gravitational force, 𝐹G, vs. distance from Earth’s

centre, 𝑟. 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 𝑟i to 𝑟f, away from the fixed point

charge 𝑄, is

𝑊 = ∫ 𝐹𝑒𝑑𝑟𝑟f

𝑟i

where 𝐹𝑒 =𝑘𝑄𝑒

𝑟2

⇒ 𝑊 = ∫𝑘𝑄𝑒

𝑟2𝑑𝑟

𝑟f

𝑟i

= [−𝑘𝑄𝑒

𝑟]𝑟i

𝑟f

= −𝑘𝑄𝑒 (1

𝑟f−

1

𝑟i)

1.5 cm

+ −

1000 V

𝐹𝑒

𝑟

𝑊

𝑟i 𝑟f

Since work is also equal to the change in potential energy,

∆𝐸P = 𝐸Pf − 𝐸Pi =−𝑘𝑄𝑒

𝑟f−

−𝑘𝑄𝑒

𝑟i

As 𝑟f → ∞, 𝐸Pf → 0 (𝐸P decreases as we move farther away from 𝑄), so

0 − 𝐸Pi =−𝑘𝑄𝑒

∞−

−𝑘𝑄𝑒

𝑟i= 0 −

−𝑘𝑄𝑒

𝑟i⇒ 𝐸Pi =

−𝑘𝑄𝑒

𝑟i

This electric potential energy will ultimately be positive: 𝑘, 𝑄, and 𝑟i are all positive, while 𝑒 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 𝑞 at a distance 𝑟 away from a point charge

source 𝑄 is given by

𝐸P =𝑘𝑄𝑞

𝑟

If 𝑄 and 𝑞 have the same charge/polarity (ie. both positive or both

negative), they will repel each other, and 𝐸P > 0.

If 𝑄 and 𝑞 have different charges/polarities (ie. one positive, one

negative), they will attract each other, and 𝐸P < 0.

As we get farther and farther away from the point charge 𝑄 (ie. 𝑟

tends to infinity), the electric potential energy approaches zero.

That is,

𝐸P∞ = 0.

Again, notice the similarities between gravitational and electric

potential energy.

From our equation for electric potential energy, we can also write

𝐸P =𝑘𝑄𝑞

𝑟= 𝑞𝑉 ∴ 𝑉 =

𝑘𝑄

𝑟

to represent the electric potential due to a point charge 𝑄.

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.

∆𝐸P = 𝑞∆𝑉 = 𝑒 (𝑘𝑄

𝑟B−

𝑘𝑄

𝑟A)

𝐸P (J)

𝑟 (m)

Same

Polarity

Opposing

Polarities

0

or ∆𝐸P = 𝐸PB − 𝐸PA = (𝑘𝑄𝑒

𝑟B) − (

𝑘𝑄𝑒

𝑟A) = 𝑘𝑄𝑒 [

1

𝑟B−

1

𝑟A]

= (9 × 109 Nm2/C2)(+5 × 10−6 C)(−1.6 × 10−19 C) [1

0.03 m−

1

0.02 m] = 1.2 × 10−13 J

Example 4: Calculate…

a) the proton’s change in potential as it moves from A to B.

𝑉A = 𝑉1A + 𝑉2A =𝑘𝑄1

𝑟1A

+𝑘𝑄2

𝑟2A

=(9 × 109 Nm2/C2)(+1 × 10−6 C)

0.09 m+

(9 × 109 Nm2/C2)(−2 × 10−6 C)

0.04 m= −3.5 × 105 V

𝑉B = 𝑉1B + 𝑉2B =𝑘𝑄1

𝑟1B

+𝑘𝑄2

𝑟2B

=(9 × 109 Nm2/C2)(+1 × 10−6 C)

0.15 m+

(9 × 109 Nm2/C2)(−2 × 10−6 C)

0.10 m= −1.2 × 105 V

∆𝑉 = 𝑉B − 𝑉A = (−1.2 × 105 V) − (−3.5 × 105 V) = 2.3 × 105 V

b) the work done to get the proton from A to B. (𝑝 = +1.6 × 10−19 C)

𝑊 = ∆𝐸P = 𝑞∆𝑉 = (+1.6 × 10−19 C)(2.3 × 105 V) = 3.7 × 10−14 J