Recent content by Efeguleroglu

Yes, I am just a poor freshman.

In a capacitor, for this formula $$ C = \epsilon \frac{A}{d}$$ dielectric constant is calculated using $$ \epsilon_{eff} = \frac{\int\epsilon dV}{V} $$ or in 2D $$ \epsilon_{eff} = \frac{\int\epsilon dA}{A} $$ I know capacitors are full of approximations but there is this formula and I don't...

Coulomb's Law $$ \vec{F} = \frac{1}{4 \pi \epsilon} \frac{q_1 q_2}{r^2} \hat{r} $$ $$ \vec{E} = \frac{1}{4 \pi \epsilon} \frac{Q}{r^2} \hat{r} $$ Let's say we want to find electric field with a distance r from charge Q. How does permittivity effects the magnitude? Will I choose the permittivity...

It is a refracted light ray. n1 and n2 are indexes. Black is object, grey is its image.

Ok I got the problem. $$a=b \\ \frac{h'}{h}=\frac{\tan\theta}{\tan\alpha}$$ That must be true. Thanks anyways.

Must object and its image be on the same line that is perpendicular to the surface? And why? Actually that's what all I need to know. If yes (1) is correct (2) is wrong, if no (1) is wrong (2) is correct.

Then $$a\neq b$$ and $$\frac{h}{h'}=\frac{n_2}{n_1}=\frac{b}{a}$$

Sorry corrected it. It is always said so. That's why I am asking.

(Black one is the object and grey is its image.) We know from Snell's Law: $$ n_1\sin\alpha=n_2\sin\theta $$ And I have been said that: $$ a=b\ (1)\\\ and\\\ \frac{h}{h'}=\frac{n_2}{n_1} \ (2) $$ Let's begin. $$...

I wrote $$a_c=\frac{T}{m}$$ But it must be equal to $$a_c=\frac{T+mgcos(\theta)}{m}$$ Actually I mistakenly used T as the centripetal force. It can easily be replaced. I just wanted to fix that.

How will I delete this thread?

That's too embarassing. I forgot to square the denominator. Forgive me for wasting your time :(

I have a pendulum and an object with radius "R" and mass "m". There are forces: constant gravitational acceleration and tension on the rope. I can write: $$x=R sin(\theta) \ \ y=R cos(\theta)$$ $$\dot{x}=R\dot{\theta}cos(\theta) \ \ \dot{y}=-R\dot{\theta}sin(\theta)$$...

I don't care my power consumption. But I got what I was chasing I think, thank you for that. It was just about definition. I constructed it on power. $$ε(t)=ε_{max}sin(\omega t)$$ $$P(t)=\frac{{ε_{max}}^2 sin^2(\omega t)}{R}$$...

So the less the inductance is, the more precise the result of the rms emf is you say.