Electric Sandwich :: Try Open Physics Challenge !

We 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 every term (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:

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

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.