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Point Charge Potential
• Physics in the news
• Coulomb’s Law and Sheets of Charge
• Constant-field vs. Point-charge Potential
• Point-charge Force, Field, PE, V expressions
• Point-charge graphical examples
• Point-charge Potential examples
• Point-charge PE examples
Physics in the news…
http://www.bloomberg.com/news/2014-03-25/malaysia-jet-traced-with-physics-in-pizza-fueledinmarsat-huddle.html
The group narrowed its initial calculations using principles of the so-called Doppler effect, named
after 19th-century Austrian physicist Christian Doppler, who explored how movement can alter a
signal profile.
The idea was to look at the velocity of the aircraft relative to the satellite. Depending on this
relative movement, the frequency received and transmitted differs from its normal value in the
same way that the sound of a passing train changes as it approaches and then moves away.
•
•
UK satellite operator used 19th century physics to trace missing plane - CNBC
http://www.cnbc.com/id/101521136
Review - Coulomb’s Law and Sheets of Charge
• Coulomb’s Law for point charge
𝐸=
𝑘𝑄
𝑟2
• In region between parallel plates
𝐸=
𝑄
𝜀𝑜 𝐴
=
𝜎
𝜀𝑜
𝑄 = 𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 𝑒𝑎𝑐ℎ 𝑝𝑙𝑎𝑡𝑒
𝜎 = 𝑐ℎ𝑎𝑟𝑔𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑛 𝑒𝑎𝑐ℎ 𝑝𝑙𝑎𝑡𝑒
𝐴 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑝𝑙𝑎𝑡𝑒
1
𝜀𝑜 = 4𝜋𝑘 = 8.85 ∙ 10−12 𝐶 2 𝑁 𝑚2
Note: now the field is constant
Coulomb’s Law and Sheets of Charge
• Coulomb’s Law + Lots of integral calculus = Constant fields
𝐸=
𝑘𝑄
𝑟2
𝑑𝑥
𝜎
𝜀𝑜
Constant-field vs. Point-charge Potential
• For constant fields.
𝐸 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑉 = 𝐸𝑑
• Point charges, must use calculus.
𝐸=
𝑉=
𝑘𝑄
𝑟2
𝑘𝑄
𝑟2
𝑑𝑟 =
𝑘𝑄
𝑟
Force, Field, PE, V for point charges
Force/PE
Force
Vector
quantities
Field
𝑞1 𝑞2
𝑭=𝑘 2
𝑟
Potential Energy
Scalar
quantities
Field/EP
𝑞1 𝑞2
𝑃𝐸 = 𝑘
𝑟
𝑞1
𝑬=𝑘 2
𝑟
Electric Potential
𝑞1
𝑉=𝑘
𝑟
Force, Field, PE, V for constant field
Force/PE
Force
Vector
quantities
Field
𝑭 = 𝑞𝑬
Potential Energy
Scalar
quantities
Field/EP
∆𝑃𝐸 = 𝑞𝐸∆𝑥
𝜎
𝑬=
𝜀𝑜
Electric Potential
∆𝑉 = 𝐸∆𝑥
Electric Potential around Point Charge
• Positive point charge
𝑉=
+𝑘𝑞
𝑟
• Negative point charge
𝑉=
−𝑘𝑞
𝑟
Point charge Electric Potential - Example 1
14. What is the electric potential 12.5 cm from a 5.00 µC point charge?
𝑉=
𝑘𝑞
𝑟
=
9∙109 𝑁𝑚2 𝐶 2 5∙10−6 𝐶
0.125 𝑚
= 3.6 ∙ 105 𝐽 𝐶
= 3.6 ∙ 105 𝑉
15. A point charge Q creates an electric potential of +113 V at a distance
of 30 cm. What is Q?
113 𝐽 𝐶 = 𝑉 =
𝑘𝑞
𝑟
=
9∙109 𝑁𝑚2 𝐶 2 q
0.3 𝑚
𝑞 = 3.77 ∙ 10−9 𝐶 = 3.77 𝑛𝐶
Point charge Electric Potential - Example 2
•
Just add as scalars. Simple!
𝑉=
=
+𝑘𝑞
𝑟
+
−𝑘𝑞
𝑟
9∙109 𝑁𝑚2 𝐶 2 20∙10−6 𝐶
0.5 𝑚
−
9∙109 𝑁𝑚2 𝐶 2 20∙10−6 𝐶
= 3.6 ∙ 105 𝑉 − 3.6 ∙ 105 𝑉 = 0𝑉
0.5 𝑚
Point charge Electric Potential - Example 3
•
Just add as scalars. Simple!
𝑉=
=
+𝑘𝑞
𝑟
+
−𝑘𝑞
𝑟
9∙109 𝑁𝑚2 𝐶 2 50∙10−6 𝐶
0.3 𝑚
+
9∙109 𝑁𝑚2 𝐶 2 −50∙10−6 𝐶
0.6 𝑚
= 1.5 ∙ 106 𝑉 − 0.75 ∙ 106 𝑉 = 7.5 ∙ 105
How does calculating E from V “work”?
• Calc potential, then field
∆𝑉
𝐸=
∆𝑥
• Calculate topo lines, then slope
∆𝑒𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛
𝑠𝑙𝑜𝑝𝑒 =
∆𝑥
Example – Problem 19
• 19
Electric Potential vs. PE
• Potential is PE without 2nd charge
𝑉=𝑘
𝑞1
𝑟
𝑃𝐸 = 𝑘
𝑞1 𝑞2
𝑟
• Electric potential around (+) charge
– Defined positive (volcanic peak)
– (+) test charge (falls away)
– (-) test charge (falls down in)
• Electric potential around (-) charge
– Defined negative (inverted peak/sinkhole)
– (+) test charge (falls down in)
– (-) test charge (falls away)
Point charge PE – Example 1
• Initially r → ∞ PE=0
• At 0.5 m separation
𝑃𝐸 =
𝑞 𝑞
𝑘 1𝑟 2
=
9∙109 𝑁𝑚2 𝐶 2 3∙10−6 𝐶 20∙10−6 𝐶
0.5 𝑚
= 1.08 𝐽
• To bring together, you have to put 1.08 J in.
• If you allow to separate, you will get 1.08 J out.
Point charge PE – Example 2
•
PE of both charges at .325 m
𝑃𝐸 =
•
𝑞 𝑞
𝑘 1𝑟 2
=
9∙109 𝑁𝑚2 𝐶 2 −0.125∙10−6 𝐶 −1.6∙10−19 𝐶
0.325 𝑚
= 5.54 ∙ 10−16 𝐽
Yes, the -0.125 μC charge recoils slightly, but we’ll ignore that. For electron, this
all goes into KE
1
2
𝑚𝑣
2
1
= 2 9.1 ∙ 10−31 𝑘𝑔 𝑣 2 = 5.54 ∙ 10−16 𝐽
𝑣 = 3.49 ∙ 107 𝑚/𝑠
Point charge PE – Example 3
• First electron requires no work
• Second electron must do work against first electron
W = ∆𝑃𝐸 =
•
𝑞 𝑞
𝑘 1 2
𝑟
10−10
𝑚
= 2.3 ∙ 10−18 𝐽
Third electron must do work against first and second electrons
W = ∆𝑃𝐸 = 𝑘
•
=
9∙109 𝑁𝑚2 𝐶 2 −1.6∙10−19 𝐶 −1.6∙10−19 𝐶
𝑞1 𝑞2
𝑟
+𝑘
𝑞1 𝑞2
𝑟
= 2.3 ∙ 10−18 𝐽 + 2.3 ∙ 10−18 𝐽
Grand total
W = 2.3 ∙ 10−18 𝐽 + 2.3 ∙ 10−18 𝐽 + 2.3 ∙ 10−18 𝐽 = 6.9 ∙ 10−18 𝐽
Point charge PE – Example 4
• Initial configuration
𝑃𝐸 =
𝑞 𝑞
𝑘 1 2
𝑟
+
𝑞 𝑞
𝑘 1 2
𝑟
=
9∙109 𝑁𝑚2 𝐶 2 35∙10−6 𝐶 0.5∙10−6 𝐶
0.16 𝑚
+
9∙109 𝑁𝑚2 𝐶 2 35∙10−6 𝐶 0.5∙10−6 𝐶
0.16 𝑚
= 1.97 𝐽
• Final configuration
𝑃𝐸 = 𝑘
𝑞1 𝑞2
𝑟
+𝑘
𝑞1 𝑞2
𝑟
=
9∙109 𝑁𝑚2 𝐶 2 35∙10−6 𝐶 0.5∙10−6 𝐶
= 4.59 𝐽
• Difference
= 4.59 𝐽 − 1.97 𝐽 = 2.62 𝐽
0.04 𝑚
+
9∙109 𝑁𝑚2 𝐶 2 35∙10−6 𝐶 0.5∙10−6 𝐶
0.24 𝑚
Potential and Field
• Field
(+) on left, (-) on right
No cancellation between charges.
Must be to right of weaker charge.
+3μC
• Potential
– (+) on left, (-) on right
– 1st Zero between charges
– 2nd Zero to right of weaker charge
-2C
Potential and Field
• Field
𝑘
3𝜇𝐶
0.05 + 𝑥
3
0.05+𝑥
−𝑘
2
2𝜇𝐶
=0
𝑥2
2
𝑥
=
+3μC
3 𝑥 = 2 .05 + 𝑥
2(0.05)
3− 2
𝑥=
= 22 𝑐𝑚 𝑡𝑜 𝑟𝑖𝑔ℎ𝑡 𝑜𝑓 − 2𝜇𝐶
• Potential
𝑘
𝑘
3𝜇𝐶
𝑥
+𝑘
3𝜇𝐶
.05+𝑥
−2𝜇𝐶
.05−𝑥
+𝑘
−2𝜇𝐶
𝑥
=0
𝑥 = 2 𝑐𝑚
=0
𝑥 = 10 𝑐𝑚
-2C
Point charge PE - Examples
• Problem 18, 19, 21, 22
Chapter 16/17
Point charge problems
𝑞1 𝑞2
𝑟2
𝑞 𝑞
𝑘 1𝑟 2
𝐹=𝑘
𝑃𝐸 =
𝑞
𝐹 = 𝑘 𝑟 21
𝑉=𝑘
𝑞1
𝑟
• Decision point
Constant field problems
𝐹 = 𝑞𝐸
𝑃𝐸 = 𝑞𝐸𝑑
𝐸
𝑉 = 𝐸𝑑