Question

A question for Physics Newton Refers to the attachment !

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Answer 1

Appropriate Question :-

A non - conducting disc of radius a and uniform positive surface charge density σ is placed on the ground , with its axis vertical . A particle of mass m and positive charge q is dropped along the axis of the disc from a height H with zero initial velocity . The particle has charge to mass ratio $\dfrac{q}{m} = \dfrac{4 gε_{0} }{σ}$ .

To Find :-

The Value of "H" , if the particle just reaches the disc .

Solution :-

At First, kindly see the attachment .

As , the Charge is at a Height "H" above the disc . So , it possesses both electrostatic potential energy and gravitational potential energy .And when the charge just reaches the disc, it possesses only electrostatic potential energy .

After Understanding the above concept , From Principle of Conservation of Energy we concluded that ;

=> Potential energy ( electrostatic + gravitational ) of the Particle at H = Electrostatic potential energy of the particle at disc

Let us assume that ;

• The Centre of the disc with surface density σ = O
• The particle of mass m and charge q is dropped from a point P.
• The Gravitational potential energy of the particle at P = $U_1$
• The Electrostatic Potential energy of the particle, at P = $U_2$
• The Electrostatic Potential due to the charged disc at point P = V

$\Rightarrow U_1 = mgH$

$\Rightarrow U_2 = Vq$

Now , In Order to Find V , Consider a small portion of the disc between two concentric circles of radii " j and j + dj " . If , dq is charge on the elementary portion , then potential at the point P due to the elementary portion is ;

$\Rightarrow dV = \dfrac{1}{4π ε_{0} } . \dfrac{dq}{r}$

As σ is the surface charge density of the disc , then ;

dq = ( 2 π j dj ) σ = 2 π σ j dj

Also,. from attachment ;

$\Rightarrow r = { ( H² + j² ) }^\dfrac{1}{2}$

{ $\because$ Pythagoras Theorem }

Now , Potential at point P due to the whole disc can be found by integrating the above within limit j = 0 to j = a ;

Integrating we get ;

$\Rightarrow V = \dfrac{σ}{2ε_{0} } . \displaystyle \int\limits_{0}^{a} { \dfrac{j} {(H² + j²)^ \dfrac{1}{2} } \, dx }$

Put , H² + j² = z

Then , 2j dj = dz

$\Rightarrow j \: dj = \dfrac{1}{2} dz$

The limits of integration corresponding to j = 0 and j = a , now become z = H² and z = H² + a² respectively ;

$\Rightarrow V = \dfrac{σ}{2ε_{0} } . \displaystyle \int\limits_{H²}^{H² + a²} { \dfrac{\dfrac{1}{2} }{(z)^\dfrac{1}{2} } \, dz }$

$\Rightarrow V = \dfrac{σ}{4ε_{0} } . \displaystyle \int\limits_{H²}^{H² + a²} { {(z)}^\dfrac{-1}{2} \, dz }$

$\Rightarrow V = \dfrac{σ}{4ε_{0} } \left|_{\:}^{\:}\dfrac{(z)^\dfrac{1}{2} }{\dfrac{1}{2} } \right |_{H²}^{H²+a²}$

On further solving we get ;

$\Rightarrow V = \dfrac{σ}{2ε_{0} } ( \sqrt{H² + a²} - H )$

Therefore , Total energy of the particle, when at Point P ,

$U = U_{1} + U_{2}$

$\Rightarrow U = mgH + \dfrac{qσ}{2ε_{0} } ( \sqrt{H² + a²} - H )$ -------( i )

The energy of the particle , when it reaches the disc , can be calculated by putting H = 0 in ( i ) . Thus , energy of the particle when it reaches the disc ;

$U' = mg.0 + \dfrac{σ}{2ε_{0} } ( \sqrt{0² + a²} - 0 )$

$\Rightarrow U' = \dfrac{qaσ}{2ε_{0} }$ -------( ii )

By the Principle of Conservation of Energy ;

$U = U'$

$\Rightarrow mgH + \dfrac{qσ}{2ε_{0} } ( \sqrt{H² + a²} - H ) = \dfrac{qaσ}{2ε_{0} }$ -----( iii )

Now , It is given to us that ;

$\dfrac{q}{m} = \dfrac{4 gε_{0} }{σ}$

$\Rightarrow \dfrac{q}{m} = \dfrac{4 g ε_{0} }{σ}$

$\Rightarrow \dfrac{m}{q} = \dfrac{σ}{4 g ε_{0} }$

$\Rightarrow m = \dfrac{qσ}{4 g ε_{0} }$

Putting $\Rightarrow m = \dfrac{qσ}{4 g ε_{0} }$ in ( iii ) we get ;

$\Rightarrow \dfrac{qσ}{4g ε_{0} } . gH + \dfrac{qσ}{2ε_{0} } ( \sqrt{ H² + a² } - H ) = \dfrac{qaσ}{2ε_{0} }$

$\Rightarrow \dfrac{H}{a} + ( \sqrt{H² + a²} - H ) = a$

$\Rightarrow \sqrt{H² + a²} = \dfrac{H}{2} + a$

On Squaring both sides we get ;

$\Rightarrow H² + a² = { ( \dfrac{H}{2} + a ) }^{2}$

$\Rightarrow H² + a² = { ( \dfrac{H}{2} ) }^{2} + a² + 2. \dfrac{H}{2} . a$

$\Rightarrow H² + a² = \dfrac{H²}{4} + a² + aH$

$\Rightarrow H² - \dfrac{H²}{4} + a² - a² = aH$

$\Rightarrow \dfrac{3H²}{4} = aH$

$\Rightarrow H = \dfrac{4}{3}a$

Henceforth , Required value of H is $\dfrac{4}{3}a$