SATHEE: Electrostatics 7 Question 29

Electrostatics 7 Question 29

31. A non-conducting disc of radius $a$ and uniform positive surface charge density $\sigma$ 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 $q / m=4 \varepsilon _0 g / \sigma$. (1999, 10M)

(a) Find the value of $H$ if the particle just reaches the disc.

(b) Sketch the potential energy of the particle as a function of its height and find its equilibrium position.

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Solution:

Potential at a height $\boldsymbol{H}$ on the axis of the disc $\boldsymbol{V}(\boldsymbol{P})$

The charge $d q$ contained in the ring shown in figure

$$ d q=(2 \pi r d r) \sigma $$

Potential at $P$ due to this ring

$$ \begin{aligned} & d V=\frac{1}{4 \pi \varepsilon _0} \cdot \frac{d q}{x} \text { where } x=\sqrt{H^{2}+r^{2}} \\ & d V=\frac{1}{4 \pi \varepsilon _0} \cdot \frac{(2 \pi r d r) \sigma}{\sqrt{H^{2}+r^{2}}}=\frac{\sigma}{2 \varepsilon _0} \frac{r d r}{\sqrt{H^{2}+r^{2}}} \end{aligned} $$

$\therefore$ Potential due to the complete disc

$$ \begin{aligned} V _p & =\int _{r=0}^{r=a} d V \\ & =\frac{\sigma}{2 \varepsilon _0} \int _{r=0}^{r=a} \frac{r d r}{\sqrt{H^{2}+r^{2}}} \\ V _p & =\frac{\sigma}{2 \varepsilon _0}\left[\sqrt{a^{2}+H^{2}}-H\right] \end{aligned} $$

Potential at centre, $(O)$ will be

$$ V _O=\frac{\sigma a}{2 \varepsilon _0} \quad H=0 $$

(a) Particle is released from $P$ and it just reaches point $O$. Therefore, from conservation of mechanical energy

Decrease in gravitational potential energy $=$ Increase in electrostatic potential energy

$$ \left(\Delta KE=0 \text { because } K _i=K _f=0\right) $$

$$ \begin{aligned} & \therefore m g H=q\left[V _O-V _p\right] \\ & \text { or } g H=\frac{q}{m} \quad \frac{\sigma}{2 \varepsilon _0}\left[a-\sqrt{a^{2}+H^{2}}+H\right] \\ & \frac{q}{m}=\frac{4 \varepsilon _0 g}{\sigma} \\ & \therefore \quad \frac{q \sigma}{2 \varepsilon _0 m}=2 g \end{aligned} $$

Substituting in Eq. (i), we get

$$ \begin{aligned} & & g H & =2 g\left[a+H-\sqrt{a^{2}+H^{2}}\right] \\ & \text { or } & \frac{H}{2} & =(a+H)-\sqrt{a^{2}+H^{2}} \\ & \text { or } & \sqrt{a^{2}+H^{2}} & =a+\frac{H}{2} \\ & \text { or } & a^{2}+H^{2} & =a^{2}+\frac{H^{2}}{4}+a H \\ & \text { or } & \frac{3}{4} H^{2} & =a H \\ & \text { or } & H & =\frac{4}{3} a \\ & \text { and } & H & =0 \\ \therefore & & H & =(4 / 3) a \end{aligned} $$

(b) Potential energy of the particle at height $H=$ Electrostatic potential energy + gravitational potential energy

$$ \therefore \quad U=q V+m g H $$

Here $V=$ Potential at height $H$

$$ U=\frac{\sigma q}{2 \varepsilon _0}\left[\sqrt{a^{2}+H^{2}}-H\right]+m g H $$

At equilibrium position $F=\frac{-d U}{d H}=0$

Differentiating Eq. (ii) w.r.t. $H$

$$ \text { or } \begin{aligned} m g+\sigma \frac{q}{2 \varepsilon _0} \frac{1}{2}(2 H) \frac{1}{\sqrt{a^{2}+H^{2}}}-1 & =0 \\ & \because \frac{\sigma q}{2 \varepsilon _0}=2 m g \end{aligned} $$

$\therefore \quad m g+2 m g \frac{H}{\sqrt{a^{2}+H^{2}}}-1=0$

or $\quad 1+\frac{2 H}{\sqrt{a^{2}+H^{2}}}-2=0$

or $\frac{2 H}{\sqrt{a^{2}+H^{2}}}=1$ or $\frac{H^{2}}{a^{2}+H^{2}}=\frac{1}{4}$

or $\quad 3 H^{2}=a^{2}$ or $H=\frac{a}{\sqrt{3}}$

From Eq. (ii), we can write

$$ U-H \text { equation as } $$