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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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:

\[2\pi \sqrt { \frac { 6m{ \varepsilon }_{ 0 } }{ \rho q(8-5\sqrt { 2 } ) } } \] – Raghav Vaidyanathan · 2 years, 3 months ago

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– Deepanshu Gupta · 2 years, 3 months ago

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 !Log in to reply

@Deepanshu Gupta @Mvs Saketh @Ronak Agarwal – Raghav Vaidyanathan · 2 years, 3 months ago

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But that doesn't seem elegant, hence I thought of differentiation method. – Ronak Agarwal · 2 years, 3 months ago

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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 :) – Mvs Saketh · 2 years, 3 months ago

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– Raghav Vaidyanathan · 2 years, 3 months ago

Okay. I actually found a lot of similarity between the method above and yours. After posting this, I tried using Newton-Leibniz rule in order to calculate the derivative of \(V(a)\) directly instead of getting \(V(a)\) and then differentiating. It does work. But I'm quite unsatisfied with the fact that this problem takes a while to solve, no matter what method one uses.Log in to reply

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. – Ronak Agarwal · 2 years, 3 months ago

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– Mvs Saketh · 2 years, 3 months ago

i see you have not used wolfram, Nice +1Log in to reply

– Deepanshu Gupta · 2 years, 3 months ago

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 ...Log in to reply

Also, i do not get notifications if some one mentions me , – Mvs Saketh · 2 years, 3 months ago

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– Deepanshu Gupta · 2 years, 3 months ago

mention and disput's are are comes on Emails only ... But okay leave that , I actually replied to your comment on My problem Fearless boy..... and where I asked something please reply there ...!Log in to reply

@Mvs Saketh – Deepanshu Gupta · 2 years, 3 months ago

did you solved it ?Log in to reply

– Mvs Saketh · 2 years, 3 months ago

i got to the expression of field for hemisphere and then thought it can be solved so i stoppedLog in to reply

– Deepanshu Gupta · 2 years, 3 months ago

Please post how much you reach .... Ronak and me discuss this earliear that this is very difficult to solve , if we use conventional method ... Roonak approach is very cool ! That's why I posted this as a note !Log in to reply

– Mvs Saketh · 2 years, 3 months ago

My approach is no different from Ronaks, i too use Newton Leibinz method to avoid Dauting Integrals or wolfram where needed, i dont think i have anything more to add to him , But i shall add if i do get some idea :)Log in to reply

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

I had puted the value R=1 . – Deepanshu Gupta · 2 years, 3 months ago

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– Raghav Vaidyanathan · 2 years, 3 months ago

I think there is an easier way to do this problem rather than finding field. Why not find potential energy? Then we can use total energy equation and differentiate it to get SHM equation.Log in to reply

– Mvs Saketh · 2 years, 3 months ago

yes you can, but whether it is really easy is not has to be tried, we do not yet know if the integral for potential of hemisphere is simpler (though it usually is the case that potential is simpler than field) (have you tried that?)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. – Raghav Vaidyanathan · 2 years, 3 months ago

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– Deepanshu Gupta · 2 years, 3 months ago

we are waiting !Log in to reply

– Raghav Vaidyanathan · 2 years, 3 months ago

It is simpler, but I think I am going wrong in some the integration somewhere. I don't think I'm getting the boundary conditions right when I substitute them in my equationLog in to reply

@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 ..! – Deepanshu Gupta · 2 years, 3 months ago

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

– Deepanshu Gupta · 2 years, 3 months ago

Have you done it with that ....?Log in to reply

But,i dont think i can solve it, (not in a simple manner surely) – Mvs Saketh · 2 years, 3 months ago

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– Raghav Vaidyanathan · 2 years, 3 months ago

I think there is an easier way to do this problem rather than finding field. Why not find potential energy? Then we can use total energy equation and differentiate it to get SHM equation.Log in to reply