[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"detail-sidebar-cat-0-en-105":3,"doc-detail-672278-en":59,"doc-seo-672278-105":79},{"code":4,"msg":5,"data":6},0,"success",[7,13,18,23,28,33,38,43,48,51,55],{"id":8,"doc_module":4,"doc_module_name":9,"category_name":10,"show_sort_weight":11,"slug":12},1,"Document","Story & Novel",90,"story-novel",{"id":14,"doc_module":4,"doc_module_name":9,"category_name":15,"show_sort_weight":16,"slug":17},2,"Literature",80,"literature",{"id":19,"doc_module":4,"doc_module_name":9,"category_name":20,"show_sort_weight":21,"slug":22},4,"Exam",70,"exam",{"id":24,"doc_module":4,"doc_module_name":9,"category_name":25,"show_sort_weight":26,"slug":27},5,"Comic",60,"comic",{"id":29,"doc_module":4,"doc_module_name":9,"category_name":30,"show_sort_weight":31,"slug":32},6,"Technology",50,"technology",{"id":34,"doc_module":4,"doc_module_name":9,"category_name":35,"show_sort_weight":36,"slug":37},7,"Healthcare",40,"healthcare",{"id":39,"doc_module":4,"doc_module_name":9,"category_name":40,"show_sort_weight":41,"slug":42},8,"Research & Report",30,"research-report",{"id":44,"doc_module":4,"doc_module_name":9,"category_name":45,"show_sort_weight":46,"slug":47},9,"Religion & Spirituality",20,"religion-spirituality",{"id":46,"doc_module":4,"doc_module_name":9,"category_name":49,"show_sort_weight":46,"slug":50},"World Cup","world-cup",{"id":52,"doc_module":4,"doc_module_name":9,"category_name":53,"show_sort_weight":52,"slug":54},10,"Lifestyle","lifestyle",{"id":56,"doc_module":4,"doc_module_name":9,"category_name":57,"show_sort_weight":24,"slug":58},19,"General","general",{"code":4,"msg":5,"data":60},{"doc_id":61,"user_id":62,"nickname":63,"user_avatar":64,"doc_module":4,"category_id":29,"category_name":30,"doc_title":65,"doc_description":66,"doc_content":67,"file_id":68,"file_url":69,"file_type":70,"file_size":71,"view_count":8,"is_deleted":4,"is_public":8,"is_downloadable":8,"audit_status":8,"page_count":14,"language":72,"language_code":73,"site_id":74,"html_lang":73,"table_of_contents":75,"faqs":76,"seo_title":77,"seo_description":66,"update_tm":78,"read_time":24},672278,1099523882367,"Jordan Avery","https://ap-avatar.wpscdn.com/davatar_9964176cb1d06d4a9deccf72a44ae3dc","PH2200 Formula Sheet - Electric Charges, Fields, and Circuits Summary","PH2200公式汇总覆盖电荷与力、电场与电势、并联与串联电容、电路与电流密度、以及磁场与电磁感应等核心物理内容。内容包含库仑力与高斯定律相关的通量表达、平行板电容与理想电池的电势差关系、串并联等效电阻/电容推导、欧姆与回路/节点定律应用。还给出安培/右手定则下磁场公式、法拉第电磁感应与麦克斯韦相关的场—波关系表达，并整理若干物理常数与几何公式。","PH2200 Formula Sheet Knight  \nElectric Charges and Forces  \nK q1 q2   \nF = F = ~~  ~~1 on 2 2 on 1 2  \nr q = ( Np − Ne ) e GE = GFon q / q  \nG G  \nFon B = qBE  \nGE = 41πε0~~ ~~qr~~ ~~2~~ ~~  point charge  \nThe Electric Field  \nGEnet = ∑i GEi Electric dipole:  \n-q  \nG  \np  \ns  \n- +  \n+q  \nGp = ( qs, from negative to positive ) Field on axis GE = ~~ ~~41πε0~~ ~~2r  \nField in bisecting plane GE =− ~~ ~~41πε0~~ ~~ linear charge density : λ = ~~ ~~QL  \nsurface charge density : η = ~~ ~~QA  \nvolume charge density : ρ = ~~ ~~QV Uniform infinite line of charge:  \nG ⎛  1 2 λ ⎞  \nE = ⎜ , perpendicular to line ⎟  \n⎝ 4πε0 r ⎠ Uniform infinite plane of charge:  \nGE = ⎛⎜ η , perpendicular to plane⎞⎟⎝ 2ε0 ⎠  \nUniformly charged sphere: GE = ~~ ~~4πQε~~ ~~0r~~ ~~2~~ ~~  for r ≥ R  \nParallel-plate capacitor:  \nG ⎛   ⎞  \nE = ⎜ η , from positive to negative ⎟⎝ ε0 ⎠  \n( Ering )z  \n( Edisk )z Ga = ( q /  \n 1 zQ   \n= 4πε0 ( z 2 + R2 )3/ 2  \n η ⎡ z ⎤= 2ε0 ⎢⎣1 −  z 2 + R2 ⎦⎥ m ) GE  \nτ = pE sinθ  \nGauss’s Law  \nΦ e = GE ⋅ GA = EAcosθ constant field  \nΦ e = ∫ GE ⋅ dGA surface  \nΦ e = GE ⋅ dGA = ~~ ~~Qε~~i~~n0~~ ~~  \nGE at surface of a charged conductor:⎧ η   \n⎪ Esurface = ⎨ ε0  \n⎪⎩perpendicular to surface  \nThe Electric Potential  \nUelect = U0 + qEs (parallel-plate capacitor) U 1 2 =  1 q1q2  Uelect = ∑  1 qiqj  q +q 4πε0 r i\u003Cj 4πε0 rij  \nUdipole =−Gp ⋅ GE =−pE cos θ Uq+sources = qV V = ~~ ~~Uq+sourcesq  \nV = Es (inside a parallel-plate capacitor ) E = ~~ ~~ΔVdC (parallel plate capacitor) V = ~~ ~~41πε0~~ ~~qr~~ ~~ point charge  \nV = ∑i 41πε0~~ ~~qrii~~ ~~  \nPotential and Field  \nΔV = V (sf ) −V(si ) = −∫ssif Es ds  \n= the negative of the \"area\"  \ndVE =− ~~  ~~s ds  \nΔVloop = ∑i ( ΔV)i = 0  \nΔVbat = ~~ ~~Wc~~h~~qem~~ ~~ = E (ideal battery)  \nC =  \nC =  \nQ  \nΔVC  \nε0 Ad  \nC = κC0  \n(parallel-plate capacitor)  \nCeq = C1 + C2 + C3 + ... (parallel)  \n⎛ 1   1 1  ⎞−1  \nCeq = ⎜ + + + ... ⎟ (series)  \n⎝ C1 C2 C3 ⎠  \nUC = ~~ ~~Q2C2~~ ~~ = ~~ ~~12 C ( ΔVC )2  \nu =  ε0  E 2  \nE  \n2  \nCurrent and Conductivity  \nElectron current:  \ni = rate of electron flow N = iΔt  \ne  \ni = nAvd  \nvd = ~~ ~~eτm~~ ~~ E Conventional current:  \nI = rate of charge flow = ei Q = I Δt  \nCurrent density: J = I / A  \nJ = nevd = σ Ene2τ 1  \nσ = ~~ ~~ = ~~  ~~m ρ  \n∑ Iin = ∑ Iout E =  ΔVwire wire  \nL  \nI = ~~ ~~Δ~~ ~~V~~wi~~reR~~ ~~ where R = ~~ ~~ρLA  \nFundamentals of Circuits  \nI = ~~ ~~ΔVR  \njunction law : ∑ Iin = ∑ Iout  \nloop law : ΔVloop = ∑i ( ΔV)i = 0 Pbat = IE  \nPR = I ΔVR = I 2R = ~~ ~~(~~ ~~Δ~~ ~~VRR~~ ~~)2  \nReq = R1 + R2 + R3 + ... + RN (series)  \n⎛ 1   1 1 1 ⎞−1  \nReq = ⎜ + + + ... + ⎟ (parallel)  \n⎝ R1 R2 R3 RN ⎠ Q = Q0 e−t /τ I = I0 e−t /τ τ = RC  \nThe Magnetic Field  \nGB =  μ0 qGv   = ⎛⎜ μ0 q v s2in θ , RHR ⎞⎟ 4π r ⎝ 4π r ⎠  \nG μ0 I ΔGs ×  ⎛ μ0I ( Δs) sinθ ⎞ B =  4π r 2  = ⎜⎝  4π r2  , RHR ⎠⎟  \nμ 0 I μ 0 NIBlong straight wire = 2π d Bcoil center = 2 R  = ( AI, from south pole to north pole )  \nGBdipole = ~~ ~~μ4π0~~ ~~2z (on axis of dipole)  GB ⋅ dGs = μoIthrough  \nBsolenoid  \nG  \nF =  \non q  \nμoNI  \n=  \nL  \nqGv × GB = ( q vB sinθ, RHR )  \nf =  qB  r =  mv cyc 2π m cyc qB  \nG G  \nFwire = IL Fparallel wires  =  × GB  \n× GB = ( ILB sinθ, RHR )  \n=  \n μ0LI1I2 2π d  \n= ( μBsinθ, RHR )  \nElectromagnetic Induction  \nE = vlB  \nΦ m = GA ⋅ GB = ABcosθ (uniform GB-field)  \nΦ m = ∫ GB ⋅ dGA area of loop  \nE = N  \n\n| dΦ\u003Cbr>per turn |\n| --- |\n| dt |\n\nEcoil = ωABNsin ωt V =  N2  V  \n2 1  \nN1  \nElectromagnetic Fields and Waves GE ⋅ dGA = ~~ ~~Qε~~i~~n0~~ ~~  \nGB ⋅ dGA = 0  \nGE ⋅ dGs = − ~~ ~~dΦdtm  \n GB ⋅ dGs = μ0Ithrough + ε0 μ0 ~~ ~~dΦdte~~  ~~GF = q ( GE + Gv × GB)  \nIdisp = ε0 ~~ ~~dΦdte~~  ~~vem = c = 1/ ε0 μ0  \nc = λf  \nE = cB  \nGS = ~~ ~~1μ0~~ ~~ ( GE × GB)  \nI =  P = S =  1 E 2 =  cε0  E 2 A avg 2cμ0 0 2 0 prad = ~~ ~~FA = Ic~~ ~~ (perfect absorber)  \nI = I0 cos2 θ  \nItransmitted = ~~ ~~12 I0 (incident light unpolarized)  \nPhysical Constants  \nK = 8.99 × 109 N ⋅ m2 /C2  \nε0 = 8.85 × 10−12 C2 /N ⋅ m2  \ne = 1.60 × 10−19 C me = 9.11 × 10−31 kg mp = 1.67 × 10−27 kg  \nμ0 = 1.26 × 10−6 ","cbCaiic4tAhpkLOx","https://ap.wps.com/l/cbCaiic4tAhpkLOx","pdf",63560,"English","en",105,"# Electric Charges and Forces\n## The Electric Field\n## The Electric Potential\n## Gauss’s Law\n# Capacitors and Current\n## Parallel-plate Capacitor\n## Current and Conductivity\n## Fundamentals of Circuits\n# The Magnetic Field\n## Magnetic Forces\n## Electromagnetic Induction\n## Electromagnetic Fields and Waves\n# Physical Constants and Useful Geometry","[{\"question\":\"电场与电势差在平行板电容中如何关联？\",\"answer\":\"文档给出平行板电容的电场与电势差关系：V（由E与板间距离或等效表达式表示），并结合UeIect与U0/qEs的形式描述电势能变化。\"},{\"question\":\"如何使用高斯定律计算电场通量？\",\"answer\":\"文档写出电场通量与积分形式：Φe 与 ∫GE·dGA 的关系，并给出带电导体表面电场方向与法向分量的表达。\"},{\"question\":\"电磁感应与电磁波之间的关键关系是什么？\",\"answer\":\"文档给出感应电动势E与磁通变化率dΦ/dt相关的式子，并在后续列出电磁场与波动关系（例如c=λf、E=cB），体现场与波的耦合。\"}]","PH2200 Formula Sheet - Electric Charges, Fields, and Circuits Summary | PDF",1793588727,{"code":4,"msg":80,"data":81},"ok",{"site_id":74,"language":73,"slug":82,"title":65,"keywords":83,"description":66,"schema_data":84,"social_meta":137,"head_meta":139,"extra_data":141,"updated_unix":78},"ph2200-formula-sheet-electric-charges-fields-and-circuits-summary","",{"@graph":85,"@context":136},[86,100,119],{"@type":87,"itemListElement":88},"BreadcrumbList",[89,93,95,98],{"item":90,"name":91,"@type":92,"position":8},"https://docshare.wps.com","Home","ListItem",{"item":94,"name":9,"@type":92,"position":14},"https://docshare.wps.com/document/",{"item":96,"name":30,"@type":92,"position":97},"https://docshare.wps.com/document/technology/",3,{"item":99,"name":65,"@type":92,"position":19},"https://docshare.wps.com/document/ph2200-formula-sheet-electric-charges-fields-and-circuits-summary/672278/",{"url":99,"name":65,"@type":101,"image":102,"author":107,"headline":65,"publisher":109,"fileFormat":112,"inLanguage":73,"description":66,"dateModified":113,"datePublished":113,"encodingFormat":112,"isAccessibleForFree":114,"interactionStatistic":115},"DigitalDocument",{"url":103,"@type":104,"width":105,"height":106},"https://docshare.wps.com/thumbnails/ph2200-formula-sheet-electric-charges-fields-and-circuits-summary/672278.png","ImageObject",300,407,{"name":63,"@type":108},"Person",{"url":90,"name":110,"@type":111},"DocShare","Organization","application/pdf","2026-11-02",true,{"@type":116,"interactionType":117,"userInteractionCount":8},"InteractionCounter",{"@type":118},"ViewAction",{"@type":120,"mainEntity":121},"FAQPage",[122,128,132],{"name":123,"@type":124,"acceptedAnswer":125},"电场与电势差在平行板电容中如何关联？","Question",{"text":126,"@type":127},"文档给出平行板电容的电场与电势差关系：V（由E与板间距离或等效表达式表示），并结合UeIect与U0/qEs的形式描述电势能变化。","Answer",{"name":129,"@type":124,"acceptedAnswer":130},"如何使用高斯定律计算电场通量？",{"text":131,"@type":127},"文档写出电场通量与积分形式：Φe 与 ∫GE·dGA 的关系，并给出带电导体表面电场方向与法向分量的表达。",{"name":133,"@type":124,"acceptedAnswer":134},"电磁感应与电磁波之间的关键关系是什么？",{"text":135,"@type":127},"文档给出感应电动势E与磁通变化率dΦ/dt相关的式子，并在后续列出电磁场与波动关系（例如c=λf、E=cB），体现场与波的耦合。","https://schema.org",{"og:url":99,"og:type":138,"og:title":65,"og:site_name":110,"og:description":66},"article",{"robots":140,"canonical":99},"index,follow",{"doc_id":61,"site_id":74}]