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# Testing paint skills( physics question)

Hi , i made this question yesterday, and it follows :

Consider an infinitely large wire (blue) having charge per unit length $$+ \lambda$$, and a finite length wire (red) of length $$l$$ and mass $$m$$ having the same charge per unit length. Its lower end rests at a height of $$h$$ units from the blue wire,as shown in the figure.

Image text

Your task is simple. Find the frequency of oscillation of red wire on being disturbed by a small distance vertically

3 years, 8 months ago

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Hi Jatin!

Here's my solution:

At equilibrium,

$$\displaystyle mg=\frac{\lambda^2}{2\pi \epsilon_0}\ln\left(\frac{h+l}{h}\right)$$

Lets displace the red wire vertically upwards by $$y$$.

The net force acting on the red wire is:

$$\displaystyle F_{net}=\frac{\lambda^2}{2\pi\epsilon_0}\ln\left(1+\frac{l}{h+y}\right)-mg$$

Using the approximation,

$$\displaystyle \ln\left(1+\frac{l}{h+y}\right)\approx \ln\left(\frac{h+l}{h}\right)-\frac{ly}{h(l+h)}$$

$$\displaystyle F_{net}=-\frac{\lambda^2ly}{2\pi h(l+h)\epsilon_0} \Rightarrow \ddot{y}=-\frac{\lambda^2l}{2m\pi h(l+h)\epsilon_0}y$$

The above equation is for SHM and the frequency can be easily deduced.

Is this correct?

I have a small question, how do you take such approximations? I had to use Wolfram Alpha, can you provide some help on this? Many thanks! :) · 3 years, 8 months ago

You should also have shown how force of interaction is $$\frac{\lambda^2}{2 \pi \epsilon_{0}} \ln\bigg(\frac{h + l}{h}\bigg)$$, I do it here,

Consider a small element of length $$dx$$ having charge $$d q = \lambda dx$$ at a distance $$x$$ from blue wire, we know that

$$E_{x} = \frac{\lambda}{2 \pi \epsilon_{0} x}$$

$$F = \displaystyle \int_{h}^{h+l} E_{x} \lambda dx = \int_{h}^{h+l} \frac{\lambda}{2 \pi \epsilon_{0} x} \lambda dx$$

= $$\frac{\lambda^2}{2 \pi \epsilon_{0}} \ln\bigg(\frac{h + l}{h}\bigg)$$ · 3 years, 8 months ago

It was correct!, but you missed $$m$$ in the denominator, i do these approximations as:

$$d(\ln (1 +\frac{l}{h})) = \frac{1}{1 + \frac{l}{h}} \times \frac{- l}{h^2} dh$$, where $$dh = y$$, $$d$$ representing small change. · 3 years, 8 months ago

Thanks Jatin! Yes, I missed the m, sorry about that, I will edit that. :)

I was wondering if you don't mind, could you please repost the same physics problem about force on mirror you posted before, I am very interested to know about its solution. Thanks! :) · 3 years, 8 months ago

It was something where we had to apply gauss law in mirror problem, but since, i was a beginner that time in making problems i later noted that it had some problem, So, i had to delete it to avoid confusion, but still i can at least e-mail it at your id, which you might tell. · 3 years, 8 months ago

Ah, I don't like to post my address in public but I remember you posted your e-mail id in one of the discussions. I remember your e-mail address so I will send you a mail. :) · 3 years, 8 months ago

How did you approximate it ?? · 3 years, 8 months ago

S I unit of force · 3 years, 3 months ago

Challenge

Solve this within a minute:

We remove the blue wire and and take the red wire to a place where, electric field varies as $$E =E_{0} e^{-x^2}$$ It stays in equilibrium at some height $$H$$ say,and i repeat the same experiment, Task remains same , find the frequency of oscillations. · 3 years, 8 months ago

lambda/root(8pie^2epsilon zero*(h+l))...I think this is the answer..if solution is needed I will provide one later..a lil' bit busy! · 3 years, 8 months ago

No, this is not the answer. · 3 years, 8 months ago