Download Phy104 - 19

Electric Potential
• Hillside analogy
• Constant electric fields
• Work and PE moving in constant field
• PE and Electric Potential in constant field
• PE and Electric Potential examples
Topographical Analogy (1)
• Consider a hillside – top and side view
• 2 quantities
– Elevation
– Slope
• 2 are related
– 𝑠𝑙𝑜𝑝𝑒 =
∆𝑒𝑙𝑒𝑣
∆𝑥
– ∆𝑒𝑙𝑒𝑣 = 𝑠𝑙𝑜𝑝𝑒 ∙ ∆𝑥
Topographical Analogy (2)
• 2 are related
– 𝑠𝑙𝑜𝑝𝑒 =
∆𝑒𝑙𝑒𝑣
∆𝑥
– ∆𝑒𝑙𝑒𝑣 = 𝑠𝑙𝑜𝑝𝑒 ∙ ∆𝑥
• Observe:
– When elevation changes
rapidly, slope is steep!
– When elevation changes
gradually, slope is
gradual!
Topographical Analogy (3)
• Another point:
– Slope can be different in
different directions (like
a vector)
–
𝑠𝑙𝑜𝑝𝑒𝑥 =
– 𝑠𝑙𝑜𝑝𝑒𝑦 =
∆𝑒𝑙𝑒𝑣
∆𝑥
∆𝑒𝑙𝑒𝑣
∆𝑦
– Elevation is just a
number (like a scalar)
Example - Mt Gretna Topo Map
Constant elevation lines
When lines are close together – its steep!
Topographical/Electrical comparison
• Elevation -> Electric Potential
– Scalar quantity V
– No direction, just number
– Units Joules/Coulomb (volts)
• Slope -> Electric Field
–
–
–
–
Vector quantity E
Components in x,y,z direction
Units Newtons/Coulomb
or Volts/m
• 𝐸𝑥 =
∆𝑉
∆𝑥
𝐸𝑦 =
∆𝑉
∆𝑦
Electric Potential and Field Map
• Highest potential near (+) charge, lowest potential near (-) charge
• Equipotential lines (green) analogous to constant elevation lines.
• Electric field lines (red) perpendicular, point straight down “hill”
Electric Field / Potential example
5.
How strong is the electric field between two parallel plates 6.0 mm apart if the
potential difference between them is 110 V?
𝑉
110 𝑉
𝑉
𝐸 = 𝑑 = 0.006 𝑚 = 18333 𝑚 = 18333
𝑁𝑚 𝐶
𝑚
𝑁
= 18333 𝐶
6. An electric field of 640 V/m is desired between two parallel plates 11.0 mm
apart. How large a voltage should be applied?
7. The electric field between 2 parallel plates connected to a 45-V battery is 1500
V/m. How far apart are the plates?
Electric Field vs Force (+/-sign)
• Given Electric Field E, how do you find force F?
𝐹 = 𝑞𝐸
Sign of charge flips direction for (+) (-) charges
(Electrons feel force “up” the hill)
-
+
Electric Potential vs Potential Energy (+/- sign)
• Given ΔV, how do you find Work (ΔPE)?
∆𝑃𝐸 = 𝑞∆𝑉
Sign of charge flips ΔPE for (+) (-) charges
(Electrons want to fall “up” the hill)
+++++++
+++++++
+
PE
∆V
-------
PE
∆V
-------
Work crossing Electric Potential - Examples
• Positive test charge
• Negative test charge
•
Change in Potential
∆𝑉 = −𝐸 ∙ ∆𝑥
•
Change in Potential Energy
∆𝑃𝐸 = 𝑞∆𝑉
Electrostatic and Gravitational analogy
• Comparison of height and gravitational PE, with electric potential and
electrostatic PE
Change in Potential
∆𝑉 = −𝐸 ∙ ∆𝑥
• At least rocks don’t fall up!
Change in Potential Energy
∆𝑃𝐸 = 𝑞∆𝑉
Example – work moving proton
• What is change in PE of a proton moving from +125 V to -55
V? What is change in PE? How much work is done?
∆𝑃𝐸 = 𝑞∆𝑉
𝐽
𝐶
= 1.6 ∙ 10−19 𝐶 (−180 )
= -2.88 ∙ 10−17
Velocity crossing Electric Potential
•
ΔPE for electron
∆𝑃𝐸 = 𝑞∆𝑉
= −1.6 ∙ 10−19 𝐶 (+5000 𝐽 𝑉)
= −8 ∙ 10−16 𝐽
•
Change in KE
1
2
𝑚𝑣
2
= −∆𝑃𝐸 = 8 ∙ 10−16 𝐽
𝑣=
2∙8∙10−16 𝐽
9.1∙10−31 𝑘𝑔
= 4.2 ∙ 107 𝑚/𝑠
Change in Potential Energy
∆𝑃𝐸 = 𝑞∆𝑉
Example problems - 1
1. How much work does the electric field do in moving a -7.5 µC charge from
ground to a point whose potential is +65 V higher?
∆𝑃𝐸 = 𝑞𝑉 = −7.5𝜇𝐶 (+65 𝐽 𝐶) = −0.00049 𝐽
𝑊 = ∆𝐾𝐸 = −∆𝑃𝐸 = 0.00049 𝐽
2. How much work does the electric field do in moving a proton from a point with a
potential of +113 V to a point where it is -45 V? Express your answer both in joules
and electron volts.
∆𝑃𝐸 = 𝑞𝑉 = 1.6 ∙ 10−19 𝐶 (−158 𝐽 𝐶) = −2.53 ∙ 10−17 𝐽
𝑊 = ∆𝐾𝐸 = −∆𝑃𝐸 = 2.53 ∙ 10−17 𝐽
2.53∙10−17 𝐽
1.6∙10−19 𝐽 𝑒𝑣
= 158 𝑒𝑣
Example problems - 2
4.
An electron acquires 3.51 ✕ 10-16 J of kinetic energy when it is
accelerated by an electric field from plate A to plate B. What is the
potential difference between the plates, and which plate is at the higher
potential?
∆𝑃𝐸 = −∆𝐾𝐸 = −3.51 ∙ 10−16 𝐽
−3.51 ∙ 10−16 𝐽 = ∆𝑃𝐸 = 𝑞𝑉 = −1.6 ∙ 10−19 𝐶 (𝑉)
𝑉=
3.51∙10−16 𝐽
1.6∙10−19 𝐶
= 2193 𝑉
Toward + electrode
Electron-volt (ev)
• Definition 1 ev is energy electron picks up going across
potential difference of 1 volt.
∆𝑃𝐸 = 𝑞𝑉 = −1.6 ∙ 10−19 𝐶 (+1 𝐽 𝐶) = −1.6 ∙ 10−19 𝐽
∆𝐾𝐸 = −∆𝑃𝐸 = 1.6 ∙ 10−19 𝐽
1 𝑒𝑣 = 1.6 ∙ 10−19 𝐽
Example problems - 3
12. What is the speed of a proton whose kinetic energy is 2.8 keV?
KE = 2.8 𝑘𝑒𝑉 = 2800 𝑒𝑣 ∙ 1.6 ∙ 10−19 𝐽 𝑒𝑣 = 4.48 ∙ 10−16 𝐽
1
1
4.48 ∙ 10−16 𝐽 = 𝐾𝐸 = 2 𝑚𝑣 2 = 2 (1.67 ∙ 10−27 𝑘𝑔)𝑣 2
𝑣 = 7.32 ∙ 105
Example problems - 4
12. The work done by an external force to move a -7.50 µC charge from
point a to point b is 25.0 ✕ 10-4 J. If the charge was started from rest and
had 4.85 ✕ 10-4 J of kinetic energy when it reached point b, what is the
magnitude of the potential difference between a and b?
𝑊𝑒𝑥𝑡 = ∆𝐾𝐸 + ∆𝑃𝐸
25 ∙ 10−4 𝐽 = 4.85 ∙ 10−4 𝐽 + ∆𝑃𝐸
∆𝑃𝐸 = 20.15 ∙ 10−4 𝐽
• Must increase PE by 20.15 ∙ 10−4 𝐽
20.15 ∙ 10−4 𝐽 = ∆𝑃𝐸 = 𝑞𝑉
20.15 ∙ 10−4 𝐽 = −7.5 𝜇𝐶 𝑉
𝑉=
20.15∙10−4 𝐽
−7.5∙10−6 𝐶
= −269 𝑉
• Must move toward negative electrode