question_answer

(i) Two conductors carrying equal and opposite charges create a non-uniform field as shown in the figure given below. What will be the capacity of this capacitor if the field along Y-axis varies as \[E=\frac{Q}{{{\varepsilon }_{0}}A}[1+B{{Y}^{2}}],\] where B is a constant.

(ii) An air capacitor has a capacitance of \[2\,\mu F,\] which becomes \[12\,\mu \,F\] when a dielectric medium is filled in the space between the plates. Find (a) dielectric constant of that material, (b) induced charge on the dielectric due to polarisation of a charge of \[6\,\mu \,C\] is given to the positive plate of the capacitor.

Or

A non-conducting disc of radius 'r' and uniform positive surface charge density \[\sigma \] is placed on the ground, with its axis along Z-axis. 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 \[q/m=4{{\varepsilon }_{0}}g/\sigma .\]

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

(ii) Draw the graph of its potential energy as a function of its height and find its equilibrium position from the centre of the disc.

Answer:

Given, \[E=\frac{Q}{{{\varepsilon }_{0}}A}[1+B{{Y}^{2}}]\] Since we know, \[E=\frac{-\,dV}{dr}\] Thus, we have \[\frac{-\,dV}{dY}=\frac{Q}{{{\varepsilon }_{0}}A}[1+B{{Y}^{2}}]\] \[\Rightarrow \]   \[-\,dV=\frac{Q}{{{\varepsilon }_{0}}A}[1+B{{Y}^{2}}]\,dY\] \[\Rightarrow \]   \[-\int_{{{V}_{A}}}^{{{V}_{B}}}{dV}=\frac{Q}{{{\varepsilon }_{0}}A}\int_{0}^{a}{[1+B{{Y}^{2}}]\,dY}\] \[\Rightarrow \]   \[{{V}_{A}}-{{V}_{B}}=\frac{Q}{{{\varepsilon }_{0}}A}{{\left[ Y+\frac{B{{Y}^{3}}}{3} \right]}^{a}}_{0}\]\[{{V}_{A}}-{{V}_{B}}=\frac{Q}{{{\varepsilon }_{0}}A}\left[ a+\frac{B{{a}^{3}}}{3} \right]=\frac{Qa}{{{\varepsilon }_{0}}A}\left[ 1+\frac{B{{a}^{2}}}{3} \right]\] \[\Rightarrow \]   \[\frac{{{V}_{A}}-{{V}_{B}}}{Q}=\frac{a}{{{\varepsilon }_{0}}A}\left[ \frac{3+B{{a}^{2}}}{3} \right]\] \[\Rightarrow \]   \[\frac{Q}{{{V}_{A}}-{{V}_{B}}}=\frac{3{{\varepsilon }_{0}}A}{a\,[3+B{{a}^{2}}]}\] Since we know, \[\frac{Q}{V}=C\] (Capacitance) Thus, the capacitance of the given conductors,                         \[C=\frac{3{{\varepsilon }_{0}}A}{a\,[3+B{{a}^{2}}]}\]             (ii) (a) According to the question, For air, the capacitance of the capacitor,    \[{{C}_{0}}=\frac{A{{\varepsilon }_{0}}}{d}=2\mu F\]             ? (i) When a dielectric medium is placed between the plates, then \[C=\frac{KA{{\varepsilon }_{0}}}{d}=12\mu F\]                       ? (ii)             Dividing Eq. (ii) by Eq. (i), we get             \[\Rightarrow \]   \[\frac{(KA{{\varepsilon }_{0}}/d)}{(A{{\varepsilon }_{0}}/d)}=\frac{12}{2}\,\,\,\,\,\,\Rightarrow \,\,\,\,\,K=6\] \[\therefore \]The value of the dielectric constant of material K is 6. (b) Induced charge due to polarisation is given by                         \[{{q}_{p}}=q\left( 1-\frac{1}{K} \right)\] where, q = charge on the capacitor \[=6\times {{10}^{-6}}C\] According to the question, Potential at P due to ring of radius x and thickness dx at height z from disc?s centre. \[dV=\frac{1}{4\pi \,\,{{\varepsilon }_{0}}}\cdot \frac{2\pi \times dx\sigma }{{{({{x}^{2}}+{{z}^{2}})}^{1/2}}}\] So, potential due to whole disc, \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\cdot 2\pi \sigma \int_{\,0}^{\,r}{\frac{xdx}{{{({{x}^{2}}+{{z}^{2}})}^{1/2}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\cdot 2\pi \sigma \int_{X=\,0}^{X=\,r}{\frac{tdt}{t}}}\]                \[[put\,\,{{x}^{2}}+{{z}^{2}}={{t}^{2}},xdx=tdt]\]            \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\cdot 2\pi \sigma \,[\sqrt{{{z}^{2}}+{{r}^{2}}-z}]\] Applying law of conservation of energy, \[mgH+\frac{1}{4\pi \,{{\varepsilon }_{0}}}\cdot \frac{2\,\pi \,\sigma \,q}{1}[\sqrt{{{H}^{2}}+{{r}^{2}}}-H]\]                         \[=\frac{1}{4\pi \,{{\varepsilon }_{0}}}\cdot 2\pi \,\sigma \,qr\] Put,      \[\frac{q}{m}=\frac{4{{\varepsilon }_{0}}g}{\sigma },\]i.e. \[q=\frac{4{{\varepsilon }_{0}}\,mg}{\sigma }\] \[mgH=\frac{\sigma }{2{{\varepsilon }_{0}}}\cdot \frac{4{{\varepsilon }_{0}}\,mg}{\sigma }[H-\sqrt{{{H}^{2}}+{{r}^{2}}}]\]                                     \[+\frac{\sigma }{2{{\varepsilon }_{0}}}\cdot \frac{4{{\varepsilon }_{0}}\,mgr}{\sigma }\] Or,        \[mg\,(H-2r)=2\,mg\,[H-\sqrt{{{H}^{2}}+{{r}^{2}}}]\] \[\therefore \]      Solving, \[H=\frac{4r}{3}\] (b) Potential energy of particle at height y,                         \[U=\sigma \frac{q}{2{{\varepsilon }_{0}}}\cdot [\sqrt{{{z}^{2}}+{{r}^{2}}}-z]+mgz\] Substituting for q, \[U=\frac{\sigma }{2{{\varepsilon }_{0}}}\frac{4{{\varepsilon }_{0}}mg}{\sigma }[\sqrt{{{z}^{2}}+{{r}^{2}}}-az]+mgy\] \[=2mg\,[\sqrt{{{z}^{2}}+{{r}^{2}}}-z]+mgz\] For equilibrium, \[\frac{dU}{dz}=0\] \[2mg\left[ \frac{12\,\,z}{2\sqrt{{{z}^{2}}+{{r}^{2}}}}-1 \right]+mg=0\] \[mg\left[ \frac{2\,z}{\sqrt{{{z}^{2}}+{{a}^{2}}}}-1 \right]+=0\]                         \[2z=\sqrt{{{z}^{2}}+{{r}^{2}}}\] \[\therefore \]      \[z=a/\sqrt{3}\] \[{{U}_{min}}=mg\left[ 2\sqrt{\frac{{{r}^{2}}}{3}+{{r}^{2}}}-\frac{2r}{\sqrt{3}}+\frac{r}{\sqrt{3}} \right]\] \[=\sqrt{3}\,mgr\]             At         \[y=0,\]\[U=2\,mgr\]             At         \[z=\frac{4r}{3},\] \[U=2\,mg\left[ \sqrt{\frac{16{{r}^{2}}}{9}+{{r}^{2}}}-\frac{4r}{3} \right]+mg\times \frac{4r}{3}\] \[=2\,mgr\] The variation in potential energy of the particle with its height from the centre of the disc.