Electrostatics AP Physics

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

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

End Slide Attraction Repulsion

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

End Slide Dipole with Water

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)

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

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

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

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?

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 = 4.50 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 = 0.88 N

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 = 4.50 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

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

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.

Electric Fields of Two Charges: End Slide Electric Fields of Two Charges:

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

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

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

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

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 = 0.1125 N FA = 0.18 N – 0.1125 N FA,B = 0.18 N to the right (on A from B) FA,C = 0.1125 N to the left (on A from C) = +0.0675 N = 0.0675 N to the right

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 = +101.25 N k qA qc (9.0x109) (6.0x10-6) (1.5x10-6) FA,C = = rA,C2 FA,B = 101.25 N to the left (on A from B) (0.030)2 FA,C = +90 N FA,C = 90 N up (on A from C)

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 = 101.25 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

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 = 101.25 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

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 = +101.25 N FA,B = 101.25 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 = +16.2 N FB,C = 16.2 N @ 36.9o N of E (on B from C)

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 = 101.25 N to the right (on B from A) FB,C = 16.2 N @ 36.9o 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

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 = 101.25 N to the right (on B from A) FB,C = 16.2 N @ 36.9o 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

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 = 101.25 N to the right (on B from A) FB,C = 16.2 N @ 36.9o 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

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