Given two identical Solid uniformly charged (\(\rho \) , \(R\)) hemisphere , initially system in equilibrium Find time period of small vertical oscillation of charged particle \(q\) and mass \(m\).

**Details** :

\( d=2m\)

\(R=1m\)

*hemisphere are fixed

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## Comments

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TopNewestWe will proceed by finding the potential at the axis of a solid hemisphere at a distance of \(a\) from its center. We can do this by first finding potential due to a disk at it's axis and then integrating to get a sphere. I am simply giving formula here, and leaving the integration as an exercise to the reader.

\(V(a)\) is the potential at distance \(a\) when \(R=1\) :

\(\phi =\frac { \rho }{ 2{ \varepsilon }_{ 0 } } \\ V(a)=\phi \left[ \frac { { a }^{ 2 } }{ 3 } -\frac { a\sqrt { { a }^{ 2 }+1 } }{ 3 } -\frac { \sqrt { { a }^{ 2 }+1 } }{ 3a } +\frac { 1 }{ 3a } +\frac { 1 }{ 2 } \right] \)

Now we know that the motion of this particle can be approximated to S.H.M. We will use the fact that total energy remains constant:

\(\text{K.E of particle+ electrostatic P.E of particle= constant}\)

\(\Rightarrow \frac { m{ v }^{ 2 } }{ 2 } +qV(a+x)+qV(a-x)=K\)

Since R.H.S is constant w.r.t time, we will differentiate to get:

\(mv\frac { dv }{ dt } +q\frac { dV(a+x) }{ dx } \times \frac { dx }{ dt } +q\frac { dV(a-x) }{ dx } \times \frac { dx }{ dt } =0\)

\(\Rightarrow m\frac { dv }{ dt } =-q\frac { dV(a+x) }{ dx }-q\frac { dV(a-x) }{ dx } \)

We can now find the R.H.S of above equation by differentiating the \(V(a)\) function.

\(\frac { dV(a+x) }{ dx } =\frac { \phi }{ 3 } \left[ -2\sqrt { { (a+x) }^{ 2 }+1 } +\frac { \sqrt { { (a }+x)^{ 2 }+1 } }{ { (a+x) }^{ 2 } } -\frac { 1 }{ { (a+x) }^{ 2 } } +2(a+x) \right] \)

In the above equation, we use the binomial approximation on

each and everyterm (even the terms under the square root) and then neglect all terms where power of \(x\) is more than \(1\). I'm skipping the simplification, and giving the final answer at \(a=1\):\(\frac { dV(a+x) }{ dx } =\frac { \phi }{ 3 } \left[ 1-\sqrt { 2 } +\frac { (8-5\sqrt { 2 } ) }{ 2 } x \right] \)

Hence: \(\frac { dV(a-x) }{ dx } =\quad -\frac { \phi }{ 3 } \left[ 1-\sqrt { 2 } -\frac { (8-5\sqrt { 2 } ) }{ 2 } x \right] \)

Therefore:

\(\Rightarrow m\frac { dv }{ dt } =-q\frac { dV(a+x) }{ dx }-q\frac { dV(a-x) }{ dx } =-\frac { \phi }{ 3 } (8-5\sqrt { 2 } )qx\)

\( m\frac { dv }{ dt }=\frac { -\rho q }{ 6{ \varepsilon }_{ 0 } } (8-5\sqrt { 2 } )qx\)

Finally, we have the expression of time period as:

\[2\pi \sqrt { \frac { 6m{ \varepsilon }_{ 0 } }{ \rho q(8-5\sqrt { 2 } ) } } \]

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@Deepanshu Gupta @Mvs Saketh @Ronak Agarwal

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I will tell you what actually happened when I tried to solve this problem, I made the integral for electric field and in the integral itself I just added a little \(x\) with \(r\) and after series of approximations calculated the result.

But that doesn't seem elegant, hence I thought of differentiation method.

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as zero order term of field is 0 (since it is equillibrium point), and 1st order is dominant and restoring, and finally second order and higher orders are neglected. So, approximation here is just as rigorious as differentiation

@Raghav Vaidyanathan - nice, atleast you confirmed the answer, and yours seems simpler though there are some complex derivatives to be done, and maybe we should expect it to be bit complicated since this is not some mathematically simple problem anyway :)

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Nice Raghav ! Now I realise finding field alway's is not good idea ... atleast for me , who know's very li'l calculus ..! You did it nicely bro ! Thanks for sharing !

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BullShit

Firstly I am using a simple result that, electric field due to a disc of radius \(r\) having a surface charge density \( \sigma \) at on point on the axis at a distance \(z\) from the centre of the disc is given by :

\[ \displaystyle E = \dfrac { \sigma }{ 2{ \varepsilon }_{ 0 } } (1-\frac { z }{ \sqrt { { z }^{ 2 }+{ r }^{ 2 } } } )\]

In my diagram I have taken a angular elemental disc whose radius is \(R\sin(\theta) \) and width of disc is given by =\(R\sin(\theta)d\theta\), also to note that here \(\sigma = \rho \times (Width) \)

Now electric field due to that element at a point that is at a distance \(r\) from the centre of the hemisphere is given by :

\[\displaystyle dE = \frac { \rho R\sin { \theta } }{ 2{ \varepsilon }_{ 0 } } (1-\frac { (r+R\cos { \theta ) } }{ \sqrt { ({ r+R\cos { \theta } ) }^{ 2 }+{ (R\sin { \theta } ) }^{ 2 } } } )d\theta \]

Integrating it our we have :

\[\displaystyle E(r) = \frac { \rho R }{ 2{ \varepsilon }_{ 0 } } \int _{ 0 }^{ \pi /2 }{ (\sin { \theta } -\frac { (r+R\cos { \theta ) } \sin { \theta } }{ \sqrt { ({ r+R\cos { \theta } ) }^{ 2 }+{ (R\sin { \theta } ) }^{ 2 } } } )d\theta } \]

Now we know that for small \(x\) we have :

\( E(r+x)=E(r)+xE'(r) \)

and

\( E(r-x)=E(r)-xE'(r) \)

We have net electric field due to both hemisphere's when particle is at a distance \(x\) from it's mean position is given by :

\( E = -2xE'(r) \)

Now by newton leibnitz rule we have :

\[\displaystyle E'(r) = \frac { \rho }{ 2{ \varepsilon }_{ 0 } } \int _{ 0 }^{ \pi /2 }{ (\frac { { R }^{ 3 }\sin ^{ 3 }{ \theta } }{ { (({ r+R\cos { \theta } ) }^{ 2 }+{ (R\sin { \theta } ) }^{ 2 }) }^{ 3/2 } } )d\theta } \]

Evaluting the derivative at \(R\) , (since the mean position is at a distance R from the centre of the two hemispheres.)

\[ \displaystyle E'(R) = \frac { \rho }{ 2{ \varepsilon }_{ 0 } } \int _{ 0 }^{ \pi /2 }{ (\sin ^{ 3 }{ \frac { \theta }{ 2 } } )d\theta } = \frac { \rho }{ 12{ \varepsilon }_{ 0 } } (8-5\sqrt { 2 } ) \]

Hence acceleration of the particle is given by :

\[ \displaystyle a = -\dfrac { \rho q }{ 6m{ \varepsilon }_{ 0 } } (8-5\sqrt { 2 } )x \]

Hence the time period is given by :

\[ \displaystyle T = 2\pi \sqrt { \dfrac { 6m{ \varepsilon }_{ 0 } }{ \rho q(8-5\sqrt { 2 } ) } } \]

I have jumped many steps it seems hence if you have any confusion ask in the comments.

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i see you have not used wolfram, Nice +1

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did you solved it ? @Mvs Saketh

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\(\displaystyle{E(r)=\cfrac { \rho }{ 2\varepsilon } { \int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ \cos { \theta } -\cfrac { (r+\sin { \theta } )\cos { \theta } }{ \sqrt { { (r) }^{ 2 }+1+2r\cos { \theta } } } d\theta } }}\)

I had puted the value R=1 .

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But,i dont think i can solve it, (not in a simple manner surely)

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@Raghav Vaidyanathan Please post what'sever you get ... may be we together complete it ... You can skips unnecceserry steps for posting here to save ur time ..!

You can post important expression here ..... The purpose of this note is exatly that ... to discuss various Possiblities !Log in to reply

@Mvs Saketh @Deepanshu Gupta @Ronak Agarwal

YesI have tried it using the potential method and have gotten the right answer. I will post an overview soon.

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I think , u are decide to not to reply me ... am i right ? :) ... I have recently mentioned you 2-3 times but you are not responding ...

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Also, i do not get notifications if some one mentions me ,

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