Ah I got it. I wasn't considering that the current was entering the circuit "outiside".
What about there's no current entering from outside? I mean, what if I have to find the total current in the circuit with the same problem data (except for i=20)?
I1 flows clockwise as well as I2. In fact I2 can't go counterclockwise from O to N, because after O we have V2 and at first there is the positive pole. But I2 follows the negative one.
The main problem is the verse of the 2 currents ## i1 ## and ## i2 ## . I think they both go clockwise, so in the node A we should have ##i1+i2=20A##. Now let's apply the second law (going clockwise). So ##ΔV1-R1*i3 - R2*i4 + ΔV2=0##, where ##i3## and ##i4## are the currents which pass through...
So for the Gaussian theorem we know that $$ \frac{Q}{e} = \vec E \cdot \vec S $$ Q's value is known so we don't need to express it as $$Q=(4/3)\pi*(R_2 ^3-R_1 ^3)*d$$ where d is the density of the charge in the volume. I've expressed the surface $$S=4\pi*x^2$$ where x is the distance of a point...
Hello everyone! I'm curiosissimo, I'm 18 and I'm an Italian high school student. In 2020/2021 I'll attend my last high school class. I'm here because some physics problems are weird and if I make a mistake, I usually waste a lot of time finding out it. But here I'll understand everything better...