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\(\displaystyle Φ=\frac{1}{24}\frac{q}{ε_0}\).
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\(\displaystyle x=2.0m\) ;\(\displaystyle −E(0)[1.5m^2]+E(2.0m)[1.5m^2]=\frac{q_{enc}}{ε_0}=−100Nm2^/C\)
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\(\displaystyle x=2.0m\);\(\displaystyle −E(0)[1.5m^2]+E(2.0m)[1.5m^2]=\frac{q_{enc}}{ε_0}=−100Nm2^/C\)