1. Electrostatics
AP Physics

2. Identify the parts of this atom Now identify the charges of each
End Slide
Parts of The Atom
Everything that you can see is made of atoms and atoms contain protons, neutrons, and electrons
Identify the parts of this atom
Proton + o
Neutron
Now identify the charges of each –
Electron

3. The Atom ? ? ? …add or remove electrons
End Slide
The Atom
Protons and Neutrons are in the center of the atom and are EXTREMELY hard to move
Electrons, however, are on the outer part of the atom and can be removed easily
So, to change the charge of an atom, you have to… ? ? ?
…add or remove electrons

4. End Slide
Attraction
Repulsion

5. Rod has a net positive charge
End Slide
Charging by induction
Electrically neutral
Charged rod is near neutral rod and charges separate
Finger touches and allows charge to leave
When finger is removed
and charged rod is removed
Rod has a
net positive charge

6. End Slide
Dipole with Water

7. Field Forces – forces that act over a distance. Coulomb’s constant (k)
End Slide
Field Forces – forces that act over a distance.
Newton’s Law of Gravity
Coulomb’s Law of Electrostatics
𝑭 𝟏,𝟐
𝑭 𝟐,𝟏
𝑭 𝟏,𝟐
𝑭 𝟐,𝟏 m1 m2 r r q1 q2
Charge
Coulombs (C)
Force (N)
m1 m2
q1 q2
Fg = G
FE = k r2 r2
Coulomb’s constant (k)
k = 9.0 x 109 Nm2/C2
Distance (m)

8. The Force is… along the line connecting the charges
End Slide
The Force is…
along the line connecting the charges
attractive if the charges are opposite
repulsive if the charges are the same

9. Inverse Square Law – force decreases by the square of the distance.
End Slide
Inverse Square Law – force decreases by the square of the distance. 1
FE µ r2
1/4 F1,2
F1,2
F1,2
F2,1
F2,1
1/4 F2,1 q1 q2 q1 q2 r 2r r
1/4 F1,2
1/4 F2,1 q1 q2 2r

10. FE = FE = FE = –2.7 N = 2.7 N attracts k q1 q2 r2
End Slide
The diagram to the right shows two metal spheres charged to +1.0 mC (+1.0x10-6 C) and –3.0 mC (–3.0x10-6 C), respectively, sitting on insulating stands separated by 10 cm (0.10 m). What is the total force between the spheres?
k q1 q2
FE = r2
(9.0x109) (1.0x10-6) (–3.0x10-6)
FE =
0.102
FE = –2.7 N
= 2.7 N attracts

11. 2.0 m
1.5 m
q1 = -4.0x10-5 C
q2 = 5.0x10-5 C
q3 = -3.0x10-5 C
The diagram above shows three metal spheres with charges of –40 mC (–4.0x10-5 C), +50 mC (+5.0x10-5 C), and –30 mC (–3.0x10-5 C), respectively. The first and second spheres are separated by 2.0 m while the second and third are separated by 1.5 m. What is the total force on each sphere?

12. End Slide
2.0 m
1.5 m
q1 = -4.0x10-5 C
q2 = 5.0x10-5 C
q3 = -3.0x10-5 C
F13
F12
k q1 q2
k q1 q3
F1,2 =
F1,3 =
r1,22
r1,32
F1,2 = N to the right
F1,3 = 0.88 N to the left
(9.0x109) (–4.0x10-5) (5.0x10-5)
(9.0x109) (–4.0x10-5) (–3.0x10-5)
F1,2 =
F1,3 =
F1 = 4.50 N – 0.88 N = 3.62 N
2.02
3.52
F1 = 3.62 N to the right
F1,2 = –4.50 N
F1,3 = N

13. F2,3 = 2.0 m 1.5 m F12 F23 k q2 q3 r2,32 F1,2 = 4.50 N to the left
End Slide
2.0 m
1.5 m
q1 = -4.0x10-5 C
q2 = 5.0x10-5 C
q3 = -3.0x10-5 C
F12
F23
k q2 q3
F2,3 =
r2,32
F1,2 = N to the left
F2,3 = 6.00 N to the right
(9.0x109) (5.0x10-5) (–3.0x10-5)
F2,3 =
F2 = 6.00 N – 4.50 N = 1.50 N
1.52
F2 = 1.50 N to the right
F2,3 = –6.00 N

14. F3 = 0.88 N – 6.00 N = –5.12 N F3 = 5.12 N to the left
End Slide
2.0 m
1.5 m
q1 = -4.0x10-5 C
q2 = 5.0x10-5 C
q3 = -3.0x10-5 C
F23
F13
F1,3 = 0.88 N to the right
F2,3 = 6.00 N to the left
F3 = 0.88 N – 6.00 N = –5.12 N
F3 = 5.12 N to the left

15. End Slide
Electric Field
The area (or field) around a charged object that can produce an electrical force on other charged objects.
Field lines compose a diagram representing direction and magnitude of the force for a positive test charge.
These lines point away from a positive charge and toward a negative charge.

16. Electric Fields of Two Charges:
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Electric Fields of Two Charges:

17. The electric field is stronger at B than at A. E
End Slide
The number of field lines starting (ending) on a positive (negative) charge is proportional to the magnitude of the charge. E
The electric field is stronger where the field lines are closer together.
The electric field is stronger at B than at A. E A
The electric field is the same for B and C. B C

18. The Millikan Oil-Drop Experiment
𝑬𝒍𝒆𝒎𝒆𝒏𝒕𝒂𝒓𝒚 𝑪𝒉𝒂𝒓𝒈𝒆 =𝟏.𝟔× 𝟏𝟎 −𝟏𝟗 𝑪

19. Electrostatic Example 1
End Slide
Electrostatic Example 1
A hydrogen atom is one proton with one electron moving around the nucleus.
What is the electrostatic force between the proton ( 𝒒 𝟏 =+𝟏.𝟔× 𝟏𝟎 −𝟏𝟗 C) and the electron ( 𝒒 𝟐 =−𝟏.𝟔× 𝟏𝟎 −𝟏𝟗 C) when the atom has a radius of 𝒓=𝟓.𝟎× 𝟏𝟎 −𝟏𝟏 m?
𝟏.𝟎× 𝟏𝟎 −𝟏𝟎 𝒎
=𝟏𝟎𝟎 𝒑𝒎
=𝟏 𝑨𝒏𝒈𝒔𝒕𝒓𝒐𝒎
k q1 q2
(9.0x109) (1.6x10-19) (-1.6x10-19)
FE = = r2
(5.0x10-11)2
𝑭 𝑬 =−𝟗.𝟐× 𝟏𝟎 −𝟖 𝑵
𝑭 𝑬 =𝟗.𝟐× 𝟏𝟎 −𝟖 𝑵 𝒐𝒇 𝒂𝒕𝒕𝒓𝒂𝒄𝒕𝒊𝒐𝒏

20. Electrostatic Example 2
End Slide
Electrostatic Example 2
A negative charge of −𝟐.𝟎× 𝟏𝟎 −𝟒 C and a positive charge of 𝟖.𝟎× 𝟏𝟎 −𝟒 C are separated by 0.30 m. What is the force between the two charges?
k q1 q2
(9.0x109) (-2.0x10-4) (8.0x10-4)
FE = = r2
(0.30)2
FE = -1.6x104 N
FE = 1.6x104 N of attraction

21. Electrostatic Example 3
End Slide
Electrostatic Example 3
Sphere A is located at the origin and has a charge of +𝟐.𝟎× 𝟏𝟎 −𝟔 C. Sphere B is located at +𝟎.𝟔𝟎 m on the x-axis and has a charge of −𝟑.𝟔× 𝟏𝟎 −𝟔 C. Sphere C is located at +𝟎.𝟖𝟎 m on the x-axis and has a charge of +𝟒.𝟎× 𝟏𝟎 −𝟔 C. Determine the net force on sphere A.
𝒒 𝑩 =−𝟑.𝟔× 𝟏𝟎 −𝟔 C
𝒒 𝑪 =+𝟒.𝟎× 𝟏𝟎 −𝟔 C
𝒒 𝑨 =+𝟐.𝟎× 𝟏𝟎 −𝟔 C
𝒙 𝑩 =𝟎.𝟔𝟎 m
𝒙 𝑪 =𝟎.𝟖𝟎 m
k qA qB
k qA qC
(9.0x109) (2.0x10-6) (-3.6x10-6)
(9.0x109) (2.0x10-6) (4.0x10-6)
FA,B =
FA,C = = =
rA,B2
rA,C2
(0.60)2
(0.80)2
FA,B = – 0.18 N
FA,C = N
FA = 0.18 N – N
FA,B = 0.18 N to the right (on A from B)
FA,C = N to the left (on A from C)
= N = N to the right

22. Electrostatic Example 4
𝒒 𝑨 =+𝟔.𝟎 𝛍𝑪
𝒓 𝑨,𝑩 =𝟒.𝟎 cm
𝒒 𝑩 =+𝟑.𝟎𝛍𝑪
𝒓 𝑨,𝑪 =𝟑.𝟎 cm
FA = ?
𝒒 𝑪 =+𝟏.𝟓 𝛍𝑪
k qA qB
(9.0x109) (6.0x10-6) (3.0x10-6)
FA,B = =
rA,B2
(0.040)2
FA,B = N
k qA qc
(9.0x109) (6.0x10-6) (1.5x10-6)
FA,C = =
rA,C2
FA,B = N to the left (on A from B)
(0.030)2
FA,C = +90 N
FA,C = 90 N up (on A from C)

23. Electrostatic Example 4
End Slide
Electrostatic Example 4
𝒒 𝑨 =+𝟔.𝟎 𝛍𝑪
𝒓 𝑨,𝑩 =𝟒.𝟎 cm
𝒒 𝑩 =+𝟑.𝟎𝛍𝑪
𝒓 𝑨,𝑪 =𝟑.𝟎 cm
FA = ?
𝒒 𝑪 =+𝟏.𝟓 𝛍𝑪
FA,B = N to the left (on A from B)
FA,C = 90 N up (on A from C)
𝑭 𝑨 = 𝟗𝟎 𝟐 + 𝟏𝟎𝟏.𝟐𝟓 𝟐
𝑭 𝑨
𝟗𝟎 𝑵
𝑭 𝑨 =𝟏𝟑𝟓.𝟓 𝑵 @
𝑵 𝒐𝒇 𝑾 𝜽
𝜽= 𝐭𝐚𝐧 −𝟏 𝟗𝟎 𝟏𝟎𝟏.𝟐𝟓
𝟏𝟎𝟏.𝟐𝟓 𝑵
Add the vectors 𝜽=
𝟒𝟏.𝟔 𝒐

24. Electrostatic Example 4: What is the force on charge on B?
End Slide
Electrostatic Example 4: What is the force on charge on B?
𝒒 𝑨 =+𝟔.𝟎 𝛍𝑪
𝒓 𝑨,𝑩 =𝟒.𝟎 cm
𝒒 𝑩 =+𝟑.𝟎𝛍𝑪
𝒓 𝑨,𝑪 =𝟑.𝟎 cm
FB = ?
FA = ?
𝒒 𝑪 =+𝟏.𝟓 𝛍𝑪
FA,B = N to the left (on A from B)
FA,C = 90 N up (on A from C)
𝑭 𝑨
𝟗𝟎 𝑵
𝑭 𝑨 =𝟏𝟑𝟓.𝟓 𝑵 @
𝟒𝟏.𝟔 𝒐
𝑵 𝒐𝒇 𝑾 𝜽
𝟏𝟎𝟏.𝟐𝟓 𝑵
Add the vectors

25. FA,B = 101.25 N to the right (on B from A) FB,C = =
Electrostatic Example 4: What is the force on charge on B?
𝒒 𝑨 =+𝟔.𝟎 𝛍𝑪
𝒓 𝑨,𝑩 =𝟒.𝟎 cm
𝐭𝐚𝐧 𝜽 = 𝟑.𝟎 𝟒.𝟎
→𝜽= 𝐭𝐚𝐧 −𝟏 𝟑.𝟎 𝟒.𝟎 𝜽 𝜽
𝒒 𝑩 =+𝟑.𝟎𝛍𝑪
𝒓 𝑨,𝑪 =𝟑.𝟎 cm
=𝟑𝟔.𝟗°
𝒓 𝑩,𝑪 = 𝟑.𝟎 𝟐 + 𝟒.𝟎 𝟐
FB = ?
𝒓 𝑩,𝑪 =𝟓.𝟎 cm
𝒒 𝑪 =+𝟏.𝟓 𝛍𝑪
FA,B = N
FA,B = N to the right (on B from A)
k qB qC
(9.0x109) (3.0x10-6) (1.5x10-6)
FB,C = =
rB,C2
(0.050)2
(????)2
FB,C = N
FB,C = o N of E (on B from C)

26. FA,B = 101.25 N to the right (on B from A)
Electrostatic Example 4: What is the force on charge on B?
𝒒 𝑨 =+𝟔.𝟎 𝛍𝑪
𝒓 𝑨,𝑩 =𝟒.𝟎 cm
𝐭𝐚𝐧 𝜽 = 𝟑.𝟎 𝟒.𝟎
→𝜽= 𝐭𝐚𝐧 −𝟏 𝟑.𝟎 𝟒.𝟎 𝜽 𝜽
𝒒 𝑩 =+𝟑.𝟎𝛍𝑪
𝒓 𝑨,𝑪 =𝟑.𝟎 cm
=𝟑𝟔.𝟗°
𝒓 𝑩,𝑪 = 𝟑.𝟎 𝟐 + 𝟒.𝟎 𝟐
FB = ?
𝒓 𝑩,𝑪 =𝟓.𝟎 cm
𝒒 𝑪 =+𝟏.𝟓 𝛍𝑪
FA,B = N to the right (on B from A)
FB,C = o N of E (on B from C)
𝑭 𝑨,𝑩
𝑭 𝑩,𝑪
𝑭 𝑩 𝒙 𝒚
𝑭 𝑨,𝑩
𝟏𝟎𝟏.𝟐𝟓𝑵
𝟏𝟔.𝟐 𝐜𝐨𝐬 𝟑𝟔.𝟗°
=𝟏𝟐.𝟗𝟓
𝟏𝟎𝟏.𝟐𝟓 𝟎
𝟏𝟏𝟒.𝟐 𝑵
𝟏𝟔.𝟐 𝑵
𝑭 𝑩,𝑪
𝟏𝟔.𝟐 𝐬𝐢𝐧 𝟑𝟔.𝟗°
=𝟗.𝟕
𝟗.𝟕 𝑵
𝟑𝟔.𝟗°
Add the vectors

27. FA,B = 101.25 N to the right (on B from A)
Electrostatic Example 4: What is the force on charge on B?
𝒒 𝑨 =+𝟔.𝟎 𝛍𝑪
𝒓 𝑨,𝑩 =𝟒.𝟎 cm
𝐭𝐚𝐧 𝜽 = 𝟑.𝟎 𝟒.𝟎
→𝜽= 𝐭𝐚𝐧 −𝟏 𝟑.𝟎 𝟒.𝟎 𝜽 𝜽
𝒒 𝑩 =+𝟑.𝟎𝛍𝑪
𝒓 𝑨,𝑪 =𝟑.𝟎 cm
=𝟑𝟔.𝟗°
𝒓 𝑩,𝑪 = 𝟑.𝟎 𝟐 + 𝟒.𝟎 𝟐
FB = ?
𝒓 𝑩,𝑪 =𝟓.𝟎 cm
𝒒 𝑪 =+𝟏.𝟓 𝛍𝑪
FA,B = N to the right (on B from A)
FB,C = o N of E (on B from C)
𝑭 𝑩
𝑭 𝑨,𝑩
𝑭 𝑩,𝑪
𝑭 𝑩 𝒙 𝒚
𝟏𝟔.𝟐 𝐜𝐨𝐬 𝟑𝟔.𝟗°
=𝟏𝟐.𝟗𝟓 𝜽
𝟏𝟎𝟏.𝟐𝟓 𝟎
𝟏𝟏𝟒.𝟐 𝑵
𝟏𝟔.𝟐 𝐬𝐢𝐧 𝟑𝟔.𝟗°
=𝟗.𝟕
𝟗.𝟕 𝑵

28. FA,B = 101.25 N to the right (on B from A)
End Slide
Electrostatic Example 4: What is the force on charge on B?
𝒒 𝑨 =+𝟔.𝟎 𝛍𝑪
𝒓 𝑨,𝑩 =𝟒.𝟎 cm
𝐭𝐚𝐧 𝜽 = 𝟑.𝟎 𝟒.𝟎
→𝜽= 𝐭𝐚𝐧 −𝟏 𝟑.𝟎 𝟒.𝟎 𝜽 𝜽
𝒒 𝑩 =+𝟑.𝟎𝛍𝑪
𝒓 𝑨,𝑪 =𝟑.𝟎 cm
=𝟑𝟔.𝟗°
𝒓 𝑩,𝑪 = 𝟑.𝟎 𝟐 + 𝟒.𝟎 𝟐
FB = ?
𝒓 𝑩,𝑪 =𝟓.𝟎 cm
𝒒 𝑪 =+𝟏.𝟓 𝛍𝑪
FA,B = N to the right (on B from A)
FB,C = o N of E (on B from C)
𝑭 𝑩
𝑭 𝑩 = 𝟏𝟏𝟒.𝟐 𝟐 + 𝟗.𝟕 𝟐
𝟗.𝟕 𝑵 𝜽
𝑭 𝑩 =𝟏𝟏𝟒.𝟔 𝑵 @
𝑵 𝒐𝒇 𝑬
𝟏𝟏𝟒.𝟐 𝑵
𝜽= 𝐭𝐚𝐧 −𝟏 𝟗.𝟕 𝟏𝟏𝟒.𝟐 𝜽=
𝟒.𝟖𝟓 𝒐

29. 𝒒 𝑨 =+𝟒.𝟎 𝒏𝑪 𝒓 𝑨,𝑩 =𝟑.𝟔 mm 𝒒 𝑩 =−𝟓.𝟏 𝒏𝑪 𝒓 𝑨,𝑪 =𝟐.𝟎 mm 𝒒 𝑪 =+𝟏.𝟓 𝒏𝑪
Electrostatic Example 4: What is the force on charge on C?
𝒒 𝑨 =+𝟒.𝟎 𝒏𝑪
𝒒 𝑩 =−𝟓.𝟏 𝒏𝑪
𝒒 𝑪 =+𝟏.𝟓 𝒏𝑪
𝒓 𝑨,𝑪 =𝟐.𝟎 mm
𝒓 𝑨,𝑩 =𝟑.𝟔 mm