1. Point Charge Potential
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

2. Review - Coulomb’s Law and Sheets of Charge
Coulomb’s Law for point charge
𝐸= 𝑘𝑄 𝑟 2
In region between parallel plates
𝐸= 𝑄 𝜀 𝑜 𝐴 = 𝜎 𝜀 𝑜
𝑄=𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 𝑒𝑎𝑐ℎ 𝑝𝑙𝑎𝑡𝑒
𝜎=𝑐ℎ𝑎𝑟𝑔𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑛 𝑒𝑎𝑐ℎ 𝑝𝑙𝑎𝑡𝑒
𝐴=𝐴𝑟𝑒𝑎 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑝𝑙𝑎𝑡𝑒
𝜀 𝑜 = 1 4𝜋𝑘 = 8.85∙ 10 −12 𝐶 2 𝑁 𝑚 2
Note: now the field is constant

3. Coulomb’s Law and Sheets of Charge
Coulomb’s Law Lots of integral calculus = Constant fields
𝐸= 𝑘𝑄 𝑟 𝑑𝑥 𝜎 𝜀 𝑜

4. Constant-field vs. Point-charge Potential
For constant fields.
𝐸=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑉=𝐸𝑑
Point charges, must use calculus.
𝐸= 𝑘𝑄 𝑟 2
𝑉= 𝑘𝑄 𝑟 2 𝑑𝑟 = 𝑘𝑄 𝑟

5. Electric Potential around Point Charge
Positive point charge
𝑉= +𝑘𝑞 𝑟
Negative point charge
𝑉= −𝑘𝑞 𝑟

6. Point charge Electric Potential - Example 1
What is the electric potential 12.5 cm from a 5.00 µC point charge?
𝑉= 𝑘𝑞 𝑟 = 9∙ 𝑁 𝑚 2 𝐶 ∙ 10 −6 𝐶 𝑚 =3.6 ∙ 𝐽 𝐶
=3.6 ∙ 𝑉
A point charge Q creates an electric potential of +113 V at a distance of 30 cm. What is Q? 𝐽 𝐶 =𝑉= 𝑘𝑞 𝑟 = 9∙ 𝑁 𝑚 2 𝐶 2 q 0.3 𝑚
𝑞=3.77∙ 10 −9 𝐶=3.77 𝑛𝐶

7. Point charge Electric Potential - Example 2
Around +20μC
𝑉= +𝑘𝑞 𝑟
= 9∙ 𝑁 𝑚 2 𝐶 ∙ 10 −6 𝐶 0.5 𝑚
=3.6∙ 𝑉
Around -20μC
𝑉= −𝑘𝑞 𝑟
= 9∙ 𝑁 𝑚 2 𝐶 2 −20∙ 10 −6 𝐶 0.5 𝑚
=−3.6∙ 𝑉

8. Point charge Electric Potential - Example 3
Just add as scalars. Simple!
𝑉= +𝑘𝑞 𝑟 + −𝑘𝑞 𝑟
= 9∙ 𝑁 𝑚 2 𝐶 ∙ 10 −6 𝐶 0.3 𝑚 + 9∙ 𝑁 𝑚 2 𝐶 2 −50∙ 10 −6 𝐶 0.6 𝑚
=1.5∙ 𝑉−0.75∙ 𝑉=7.5∙ 10 5

9. How does calculating E from V “work”?
Calc potential, then field
𝐸= ∆𝑉 ∆𝑥
Calculate topo lines, then slope
𝑠𝑙𝑜𝑝𝑒= ∆𝑒𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛 ∆𝑥

10. Example – Problem 22 𝑉= 𝑘(+3𝑄) 𝐿 + 𝐾(−2𝑄) 𝐿 + 𝑘(𝑄) 2 𝐿
𝑉= 𝑘(+3𝑄) 𝐿 + 𝐾(−2𝑄) 𝐿 + 𝑘(𝑄) 2 𝐿
Just add as scalars, no X-Y table
(Physicists always calculate V first, then 𝐸=− 𝜕𝑉 𝜕𝑥 , etc)

11. Electric Potential vs. PE
Potential is PE without 2nd charge
𝑉=𝑘 𝑞 1 𝑟 𝑃𝐸=𝑘 𝑞 1 𝑞 2 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) + - PE + - PE

12. Point charge PE – Example 1
Initially r → ∞ PE=0
At 0.5 m separation
𝑃𝐸=𝑘 𝑞 1 𝑞 2 𝑟 = 9∙ 𝑁 𝑚 2 𝐶 ∙ 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 get 1.08 J out.

13. Point charge PE – Example 2
PE of both charges at .325 m
𝑃𝐸=𝑘 𝑞 1 𝑞 2 𝑟 = 9∙ 𝑁 𝑚 2 𝐶 2 −0.125∙ 10 −6 𝐶 −1.6∙ 10 −19 𝐶 𝑚 =5.54∙ 10 −16 𝐽
Yes, the μC charge recoils slightly, but we’ll ignore that. For electron, this all goes into KE
1 2 𝑚 𝑣 2 = ∙ 10 −31 𝑘𝑔 𝑣 2 =5.54∙ 10 −16 𝐽
𝑣=3.49∙ 𝑚/𝑠

14. Point charge PE – Example 3
First electron requires no work
Second electron must do work against first electron
W=∆𝑃𝐸=𝑘 𝑞 1 𝑞 2 𝑟 = 9∙ 𝑁 𝑚 2 𝐶 2 −1.6∙ 10 −19 𝐶 −1.6∙ 10 −19 𝐶 10 −10 𝑚 =2.3∙ 10 −18 𝐽
Third electron must do work against first and second electrons
W=∆𝑃𝐸=𝑘 𝑞 1 𝑞 2 𝑟 +𝑘 𝑞 1 𝑞 2 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

15. Point charge PE – Example 4
Initial configuration
𝑃𝐸=𝑘 𝑞 1 𝑞 2 𝑟 +𝑘 𝑞 1 𝑞 2 𝑟 = 9∙ 𝑁 𝑚 2 𝐶 ∙ 10 −6 𝐶 0.5∙ 10 −6 𝐶 𝑚 + 9∙ 𝑁 𝑚 2 𝐶 ∙ 10 −6 𝐶 0.5∙ 10 −6 𝐶 𝑚
=1.97 𝐽
Final configuration
𝑃𝐸=𝑘 𝑞 1 𝑞 2 𝑟 +𝑘 𝑞 1 𝑞 2 𝑟 = 9∙ 𝑁 𝑚 2 𝐶 ∙ 10 −6 𝐶 0.5∙ 10 −6 𝐶 𝑚 + 9∙ 𝑁 𝑚 2 𝐶 ∙ 10 −6 𝐶 0.5∙ 10 −6 𝐶 𝑚
=4.50 𝐽
Difference
=4.50 𝐽 −1.97 𝐽=2.53 𝐽

16. Potential and Field Field Potential +3μC -2C (+) on left, (-) on right
No cancellation between charges.
Must be to right of weaker charge.
Potential
1st Zero between charges
2nd Zero to right of weaker charge
+3μC C

17. Potential and Field Field 3 𝑥= 2 .05+𝑥 Potential
𝑘 3𝜇𝐶 𝑥 2 −𝑘 2𝜇𝐶 𝑥 2 =0
𝑥 = 2 𝑥
3 𝑥= 𝑥
𝑥= 2 (0.05) 3 − 2 =22 𝑐𝑚 𝑡𝑜 𝑟𝑖𝑔ℎ𝑡 𝑜𝑓 −2𝜇𝐶
Potential
𝑘 3𝜇𝐶 𝑥 +𝑘 −2𝜇𝐶 .05−𝑥 = 𝑥=3 𝑐𝑚
𝑘 3𝜇𝐶 .05+𝑥 +𝑘 −2𝜇𝐶 𝑥 = 𝑥=10 𝑐𝑚
+3μC C

18. Summary Chapter 16/17 Point charge problems
𝐹=𝑘 𝑞 1 𝑞 2 𝑟 𝐹=𝑘 𝑞 1 𝑟 2
𝑃𝐸=𝑘 𝑞 1 𝑞 2 𝑟 𝑉=𝑘 𝑞 1 𝑟
Decision point
Constant field problems
𝐹=𝑞𝐸 𝐸
𝑃𝐸=𝑞𝐸𝑑 𝑉=𝐸𝑑

19. Point charge PE - Examples
Problem 21, 24,

20. Force, Field, PE, V for point charges
Force/PE
Field/EP
Vector quantities
Force
𝑭=𝑘 𝑞 1 𝑞 2 𝑟 2
Field
𝑬=𝑘 𝑞 1 𝑟 2
Scalar quantities
Potential Energy
𝑃𝐸=𝑘 𝑞 1 𝑞 2 𝑟
Electric Potential
𝑉=𝑘 𝑞 1 𝑟

21. Force, Field, PE, V for constant field
Force/PE
Field/EP
Vector quantities
Force
𝑭=𝑞𝑬
Field
𝑬= 𝜎 𝜀 𝑜
Scalar quantities
Potential Energy
∆𝑃𝐸=𝑞𝐸∆𝑥
Electric Potential
∆𝑉=𝐸∆𝑥