TAILIEUCHUNG - Ebook Griffiths intro to electrodynamics (3rd edition): Part 1

(BQ) Part 3 combines both in a logical fashion by considering simple cases and working on up to the holy grail equation of the force on one particle by another particle (which is so obsurdly complicated nobody uses it). Later we learn Part 1 and Part 2 were the same because magnetism and electrostatics are really the same phenomena observed in different frames thanks to relativity. | Chapter 7 Electrodynamics Problem a Let Q be the charge on the inner shell. Then E -Ị-Ỵ-ẬÍ in the space between them and vo H r a E dr Q dr -rQ L 1 . Jb 47TÉ0 Jo r2 47T o ữ bJ I IJ da ơ ị E da Ơ Ơ 47 e0 K - vb e0 ì a-1 ố -Vb 4-707 . - V l a-1 ò 1 1 1 4-7 7 a b J b c For large b the second term is negligible and R 1 4-7 cm. Essentially all of the resistance is in the region right around the inner sphere. Successive shells as you go out contribute less and less because the cross-sectional area 47 r2 gets larger and larger. For the two submerged spheres R one R as the current leaves the first one R as it converges on the second . Therefore I V R I 2 ìĩơaV. I Problem a V Q C IR. Because positive I means the charge on the capacitor is decreasing -I -- ộ so Q t Qoe-t J at Hence 7 t - CVo e 0 at LIL . But Qo Q 0 CVo so Q t CVoe t RC. Ỵữ_ t RC R rco t 2 -0O ___. Pdt I2Rdt -ị e 2t RCdt _ Jo R Jo Iff i RC -2t Rc _ 1 7 2 z R V 2 0 2 v c Vo Q C IR. This time positive I means Q is increasing I - CVo - Q - at RCJ Q c Vo -I- constant Q t CVo ke t lỉG. But Q 0 0 k -CVo so rtC i CVo - H dt RU b w The energy delivered to the resistor is - dt In Q-CVo ztC Q i CVo 1 - . Là Í RC R 125 126 CHAPTER 7. ELECTRODYNAMICS T 2 H The final energy in roo VĨ I03 VĨ Í d Energy from battery Vol dt --T e t RCdt --T RCeE1 11 Jo R Jo R A Since t is the same as in a the energy delivered to the resistor is again so I half I the energy from the batter goes to the capacitor and the other half icv02. cv2. the capacitor is also r5-Cl Q2 to the Problem a I J J da where the integral is taken over a surface enclosing the positively charged conductor. But J ffE and Gauss s law says fE da Yq so I Ơ JE r da But Q cv and V IR so qed I -J-CIR or B ơC q .cv cir -i - time constant is T RC o 7. rqQ Q t Qoe t fiC or since V Q C v t voe-t 7ic. The Problem I J s 2tĩsL J s I 2tĩsL. E J ơ I 2tĩsơL I 27ĩkL. V - E-dl --4- a- . So B 7 2T kL 1 b a 2 kL Problem ị2r . 82 r R 2 dR 2R Problem

• Điện - Điện tử
• Griffiths intro to electrodynamics
• Vector analysis
• Special techniques
• Electrostatic fields
• Electrostatic fields in matter
• Magnetostatic fields in matter
• Magnetostatic fields
• Conservation laws
• Electromagnetic waves
• Potentials and fields
• Electrodynamics and relativity
• Griffiths intro
• Electric fields in matter
• magnetic fields in matter