Physics 16/21 electromagnetism Electric Charge and Electric Force MARLON FLORES SACEDON

electromagnetism What is electromagnetism? A branch of physics which involves the study of the electromagnetic force, a type of physical interaction that occurs between electrically charged particles. The electromagnetic force usually shows electromagnetic fields, such as electric fields, magnetic fields, and light.

Law of electric charges Two positive charges or two negative charges repel each other. 2. A positive charge and a negative charge attract each other. 3. Uncharge material is always attracted by a charge material. + + + + + + + + - - - - - - - - - + - + + - + - + + + +

Law of electric charges

conductors, insuLators, and induced charges

electric charge and the structure of Matter Mass of electron = me = 9.11x10-31 kg Mass of proton = mp = 1.673x10-27 kg Mass of neutron = mn = 1.675x10-27 kg Charge of One electron: e=1.602𝑥 10 −19 𝐶/electron Principle of conservation of charge: The algebraic sum of all the electric charges in any closed system is constant.

Atomic number 𝑚 𝑡𝑜𝑡 =nM 𝑁=𝑛 𝑁 𝐴 Where: 𝑚 𝑡𝑜𝑡 = total mass (g) 𝑛 = number of moles (mol) 𝑀 = molar mass (g/mol) 𝑁 𝐴 = 6.022 x1023 atoms/mol Abogadro’s number 𝑁 = number of atom in a substance 𝑚 𝑡𝑜𝑡 =nM 𝑁=𝑛 𝑁 𝐴

Coulomb’s law Coulomb's law, or Coulomb's inverse-square law, is a law of physics describing the electrostatic interaction between electrically charged particles. The law was first published in 1785 by French physicist Charles Augustin de Coulomb and was essential to the development of the theory of electromagnetism. Coulomb’s torsion balance

Coulomb’s law 𝐹 =𝑘 𝑞 1 𝑞 2 𝑟 2 𝑟 Electric force vector 𝑞 2 𝐹 12 - 𝐹 21 + 𝑞 1 𝐹= 1 4𝜋 𝜖 𝑜 ∙ 𝑞 1 𝑞 2 𝑟 2 𝐹∝ 𝑞 1 𝑞 2 𝑟 2 𝐹=𝑘 𝑞 1 𝑞 2 𝑟 2 Magnitude of electric force 𝑞 2 𝐹 =𝑘 𝑞 1 𝑞 2 𝑟 2 𝑟 Electric force vector 𝐹 12 𝐹 21 𝐹 12 + 𝑞 1 =+25𝑛𝐶 - 𝑞 2 =−75𝑛𝐶 𝑟=3.0𝑐𝑚 + 𝑟 𝑞 1 Where: 𝑘 is electric constant + 𝐹 21 (Coulomb’s law: force between two point charges) 𝑞 1 𝑎𝑛𝑑 𝑞 2 are interacting charges, in 𝐶𝑜𝑢𝑙 𝑜𝑟 𝐶 𝐹 12 =9𝑥 10 9 𝑁∙ 𝑚 2 𝐶 2 +25𝑛𝐶 75𝑛𝐶 3𝑐𝑚 2 1 coul = 2.997 x 109 statC 𝜖 𝑜 =8.854𝑥 10 −12 𝐶 2 𝑁∙ 𝑚 2 𝐹 12 =9𝑥 10 9 𝑁∙ 𝑚 2 𝐶 2 +25𝑥 10 −9 𝐶 75𝑥 10 −9 𝐶 3𝑥 10 −2 𝑚 2 Permittivity of free space Problem: Two point charges, 𝑞 1 =+25𝑛𝐶 and 𝑞 2 =−75𝑛𝐶, are separated by a distance 𝑟=3.0𝑐𝑚. Find the magnitude of the electric force (a) that 𝑞 1 exerts on 𝑞 2 and (b) that 𝑞 2 exerts on 𝑞 1 . 1 4𝜋 𝜖 𝑜 =𝑘=8.988𝑥 10 9 𝑁∙ 𝑚 2 𝐶 2 𝐹 12 =0.01875𝑁 attractive force 𝐹 21 =9𝑥 10 9 𝑁∙ 𝑚 2 𝐶 2 75𝑥 10 −9 𝐶 +25𝑥 10 −9 𝐶 3𝑥 10 −2 𝑚 2 𝑘=9𝑥 10 9 𝑁∙ 𝑚 2 𝐶 2 𝑘=1 𝑑𝑦𝑛𝑒∙𝑐 𝑚 2 𝑠𝑡𝑎𝑡𝐶 2 𝐹 21 =0.01875𝑁 attractive force

Coulomb’s law Superposition of electric charge 𝑦 𝑥 𝐹= 1 4𝜋 𝜖 𝑜 ∙ 𝑞 1 𝑞 2 𝑟 2 - −𝑞 3 𝑟 13 𝐹 21 = 1 4𝜋 𝜖 𝑜 ∙ 𝑞 1 𝑞 2 𝑟 12 2 + 𝑞 2 𝐹 31 = 1 4𝜋 𝜖 𝑜 ∙ 𝑞 1 𝑞 3 𝑟 13 2 𝐹 𝐹 31 𝑟 12 + 𝑞 1 𝐹 21 𝐹 = 𝐹 21 + 𝐹 31 Note: vector sum or geometric sum not arithmetic sum

Coulomb’s law Example. 𝐹 = 𝐹 21 + 𝐹 31 = 𝐹 𝑥 𝑖 + 𝐹 𝑦 𝑗 Calculate the net electric force on charge 1 due to charge 2 & charge 3. 𝑦 𝑥 𝐹= 1 4𝜋 𝜖 𝑜 ∙ 𝑞 1 𝑞 2 𝑟 2 1 4𝜋 𝜖 𝑜 =𝑘=9𝑥 10 9 𝑁∙ 𝑚 2 𝐶 2 + 𝑞 3 =7𝑛𝐶 3cm 𝐹 21 =9𝑥 10 9 𝑁∙ 𝑚 2 𝐶 2 ∙ (3𝑛𝐶)(9𝑛𝐶) (6𝑐𝑚) 2 𝑟 13 = 2 3 2 + 4 2 =5𝑐𝑚 𝑟 13 4cm 𝜃= 𝑇𝑎𝑛 −1 4 3 = 53.1 𝑜 =9𝑥 10 9 𝑁∙ 𝑚 2 𝐶 2 ∙ 3𝑥 10 −9 𝐶 9𝑥 10 −9 𝐶 6𝑥 10 −2 𝑚 2 =6.75𝑥 10 −5 𝑁 + 𝑞 1 =3𝑛𝐶 𝜃 𝑟 12 - 𝑞 2 =−9𝑛𝐶 𝐹 31 =9𝑥 10 9 𝑁∙ 𝑚 2 𝐶 2 ∙ (3𝑥 10 −9 𝐶)(7𝑥 10 −9 𝐶) (5𝑥 10 −2 𝑚) 2 𝜃 𝛽 =7.56𝑥 10 −5 𝑁 6cm 𝐹 31 𝐹 31 𝐹 = ? Using component method: =2.21𝑥 10 −5 𝑁 Σ 𝐹 𝑥 =6.75𝑥 10 −5 𝐶−7.56𝑥 10 −5 𝐶(cos 53.1 𝑜 ) =−6.06𝑥 10 −5 𝑁 Σ 𝐹 𝑦 =0−7.56𝑥 10 −5 𝐶(sin 53.1 𝑜 ) vector sum 𝐹= (2.21𝑥 10 −5 ) 2 + (6.06𝑥 10 −5 ) 2 =6.45𝑥 10 −5 𝑁 𝐹 = 𝐹 21 + 𝐹 31 = 𝐹 𝑥 𝑖 + 𝐹 𝑦 𝑗 Magnitude 𝛽= 𝑇𝑎𝑛 −1 Σ 𝐹 𝑦 Σ 𝐹 𝑥 = 𝑇𝑎𝑛 −1 6.06𝑥 10 −5 2.21𝑥 10 −5 𝐹 =(2.21𝑥 10 −5 𝑁) 𝑖 +(−6.06𝑥 10 −5 𝑁) 𝑗 𝐹 = 2.21𝑥 10 −5 𝑁 𝑖 − 6.06𝑥 10 −5 𝑁 𝑗 ANSWER = 70 𝑜 from +x-axis in QIV Direction

Coulomb’s law Assignment 1 Assignment 2 Two point charges are located on the +x-axis of a coordinate system: 𝑞 1 =1.0𝑛𝐶 is at 𝑥=+2.0𝑐𝑚, and 𝑞 2 =−3.0𝑛𝐶 is at 𝑥=+4.0𝑐𝑚. What is the total electric force exerted by 𝑞 1 and 𝑞 2 on a charge 𝑞 3 =5.0 𝑛𝐶 at 𝑥=0? Assignment 2 Calculate the net electric force on charge 1 due to charge 2 & charge 3. Calculate also the magnitude and direction of the net force. 𝑦 𝑥 - 𝑞 4 =4𝜇𝐶 + 𝑞 3 =−6𝜇𝐶 5cm 4cm 𝑟 13 6cm 6cm + 𝑞 1 =3𝜇𝐶 𝜃 𝑟 12 - 𝑞 2 =9𝜇𝐶 𝜃 12cm

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