Loading [MathJax]/extensions/TeX/color.js
Skip to main content
Library homepage
 
Global

Search

Searching in
About 3 results
  • https://query.libretexts.org/Kiswahili/Chuo_Kikuu_Fizikia_II_-_Thermodynamics%2C_Umeme%2C_na_Magnetism_(OpenStax)/06%3A_Sheria_ya_Gauss/6.0A%3A_Sheria_ya_Gauss_(Majibu)
    Kufanya mchemraba na q katika kituo, kwa kutumia mchemraba wa upande a Hii itachukua cubes nne za upande a kufanya upande mmoja wa mchemraba kubwa Upande wa kivuli wa mchemraba mdogo ungekuwa 1/24 ya ...Kufanya mchemraba na q katika kituo, kwa kutumia mchemraba wa upande a Hii itachukua cubes nne za upande a kufanya upande mmoja wa mchemraba kubwa Upande wa kivuli wa mchemraba mdogo ungekuwa 1/24 ya jumla ya eneo la mchemraba mkubwa; kwa hiyo, mtiririko kupitia eneo la kivuli ungekuwa\(\displaystyle Φ=\frac{1}{24}\frac{q}{ε_0}\).
  • https://query.libretexts.org/Francais/Physique_universitaire_II_-_Thermodynamique%2C_%C3%A9lectricit%C3%A9_et_magn%C3%A9tisme_(OpenStax)/06%3A_La_loi_de_Gauss/6.0A%3A_La_loi_de_Gauss_(r%C3%A9ponses)
    77. \(\displaystyle Φ=\frac{q_{enc}}{ε_0}\); Il y a deux contributions à l'intégrale de surface : l'une sur le côté du rectangle\(\displaystyle x=0\) et l'autre sur le côté à\(\displaystyle x=2.0m\) ;...77. \(\displaystyle Φ=\frac{q_{enc}}{ε_0}\); Il y a deux contributions à l'intégrale de surface : l'une sur le côté du rectangle\(\displaystyle x=0\) et l'autre sur le côté à\(\displaystyle x=2.0m\) ;\(\displaystyle −E(0)[1.5m^2]+E(2.0m)[1.5m^2]=\frac{q_{enc}}{ε_0}=−100Nm2^/C\)
  • https://query.libretexts.org/Idioma_Portugues/Fisica_Universitaria_II_-_Termodinamica_Eletricidade_e_Magnetismo_(OpenStax)/06%3A_Lei_de_Gauss/6.0A%3A_Lei_de_Gauss_(Respostas)
    77. \(\displaystyle Φ=\frac{q_{enc}}{ε_0}\); Existem duas contribuições para a integral da superfície: uma na lateral do retângulo em\(\displaystyle x=0\) e a outra no lado em\(\displaystyle x=2.0m\);...77. \(\displaystyle Φ=\frac{q_{enc}}{ε_0}\); Existem duas contribuições para a integral da superfície: uma na lateral do retângulo em\(\displaystyle x=0\) e a outra no lado em\(\displaystyle x=2.0m\);\(\displaystyle −E(0)[1.5m^2]+E(2.0m)[1.5m^2]=\frac{q_{enc}}{ε_0}=−100Nm2^/C\)