Recent content by MarcusNTran

ahhh I finally understand ^^.Thanks a lot for the help, really appreciate you taking the time to answer. Have a nice day :)

hmmm, I am not sure if i understand why is it that i can not reduce the intergral to EA when it comes to a sylinder? could you elaborate would help me quite a lot in the upcoming exam ^^

I moved the intergral because when we multiply the electric field to the area of element which is perpendicular to each other and the electric field is the same anywhere along the sphere then it becomes a constant that's why its become E times the intergral of dA.

It is the total flux in that point? or it is perpendicular to the sphere?

hmmm, what kind of important characteristic is it?

the electric field line would be perpendicular to the surface of the sphere, and since we multiply electric field and the area element and since theyre perpendicular to each other then it can be written as E*intergral dA=Q/e and the intergral dA is the area that its replaced with 4*pi*r^2...

tbh i only used a sphererical gaussian surface mainly because it said i should use it on my textbook. and i filled in the formula, but it just seemed abit weird that the exam had the same answer in both of problems. typo meant spherical

meant 4πr^2, so the answer in a and b is E=0? I used a sylinder as the closed surface

I used this formula, simplified the formula to become E*A=Q/E the A replaced with 2*Pi*r^2 but in both problem a and b, i found out that Q becomes zero. due to if it is hollow there will be no electrical field inside the hollow object and in the other problem q becomes zero because qa and qb...

Homework Statement Two conducting hollow spheres are are placed concentrically, the inner sphere have a radius ra = 5 cm and the outer sphere have a radius rb = 15 cm. The charge on the inner sphere is qa = 4 · 10−7 C and qb = −4 · 10−7 on the outer sphere. (a) Use Gauss’s law to find the...