Recent content by ananonanunes

This is very helpflul, I would have never noticed the mistake by myself. Thanks for pointing it out

Okay, that makes sense. Thank you so much for your help!

Okay so for a parallel plate capacitor I have ##C=\frac{\epsilon_0 A}{d}## so the final capacitance should be ##C_f=\frac{\epsilon_0 A}{d(1-f)}=\frac{C}{1-f}##. If I write ##q=\frac{C}{1-f}\Delta V## I would get ##k=\frac{C(\Delta V)^2}{d^2f(1-f)^2}## which is still not the correct answer but...

I think the capacitance depends on the distance between the plates, so no. Would this mean my mistake was assuming ##q=C\Delta V##?

This circuit consis of two identical, parallel metal plates free to move, other than being connected to identical metal springs, a switch, and a battery with terminal voltage ##\Delta V##. With the switch open, the plates are uncharged, are separated by a distance ##d##, and have a capacitance...

I understand what you mean, thanks for the help

I think that might be where my problem was but now that I look at it again it should be something like 0≤r≤2sinθ ; 0≤θ≤π

I tried integrating first z then r then teta because i wrote z in terms of r and r in terms of teta

I am given this expression which represents an object in 3D and the goal is to determine its volume using multiple integrals. I started by drawing what I think is the object as well as two "slices" of that object on different planes (z=2 and z=1) I have tried using cartesian, cylindrical and...

I just calculated the voltage and the charge using that voltage and I got the correct result. Thank you so much for the help!

Since there is no current through the ##3k\Omega## there is also no voltage across that resitor. So I thought the voltage across the capacitor would be the same as the electromotive force, because the voltage wouldn't need to be "split" between the capacitor and the resistor. But this is...

Suppose the switch has been closed for a long time so that the capacitor is fully charged and current is constant. a)Find the current in each resistor and charge Q of the capacitor. b)The switch is now opened at t=0s. Write the equation for the current for the resistor of 15kΩ as a function of...

It does. So if I have a two variable function and I want to calculate the limit for (x,y) to (0,0) I can just replace x and y with the polar coordinates and check whether it depends on \theta and if it does, there is no limit?

You're right! I totally forgot about the numbers between 0 and 1. Thanks for the help

This is what I did: $$\lim_ {(x,y) \rightarrow (1,0)} {\frac {g(x)(x-1)^2y}{2(x-1)^4+y^2}}=\lim_ {(x,y) \rightarrow (1,0)} {g(x)y\frac {(x-1)^2}{2(x-1)^4+y^2}}$$ I know that ##\lim_ {(x,y) \rightarrow (1,0)} {g(x)y}=0## and that ##\frac {(x-1)^2}{2(x-1)^4+y^2}## is limited because ##0\leq...