Electrostatics Practice Questions

Now ${Q}_{1}+ {Q}_{2}= {Q}_{1}'+ {Q}_{2}'$ $=12 \mu{C}-3 \mu{C}=9 \mu{C}$ $\& \ {V}_{1}= {V}_{2}$ $\Rightarrow {{KQ}_{1}'}/{ \frac{2 {R}}{3}}={{KQ}_{2}'}/{\frac{{R}}{3}}$ $Q_{1}'=2 Q_{2}'$ $\Rightarrow 2 Q_{2}'+Q_{2}'=9 \mu C$ $\Rightarrow Q_{2}'=3 \mu C$ $\& \ {Q}_{1}'=6 \mu{C}$

$\frac{{kQq}}{ {R}}+ {mgy}$ $=\frac{{kQq}}{ {R}+ {y}}+ \frac{1}{2} {mv}^{2}$ ${v}^{2}=2 {gy}+ \frac{2 {kQqy}}{ {mR}( {R}+ {y})}$

Let particle have charge $q$ and mass 'm'Solve for $( {q}, {m})$ mathematically ${F}_{ {x}}=0, {a}_{ {x}}=0,( {v})_{ {x}}= {\text{constant}}$ time taken to reach at 'P' $=\frac{{d}}{ {v}_{0}}= {t}_{0}( {\text{let}})$ ...... (1) (Along$- {y}), {y}_{0}=0+ \frac{1}{2} .\frac{ {qE}}{ {m}}$ $.{t}_{0}^{2}$ .... .(2) ${v}_{ {x}}= {v}_{0}$ ${v}= {u}+ {at}$ (along -ve 'y') speed $v_{y 0}= \frac{q E}{m} .t_{0}$ $\tan \theta = \frac{v_{y}}{v_{x}}= \frac{q E t_{0}}{m \.v_{0}},$ $(t_{0}= \frac{d}{v_{0}} )$ $\tan \theta =\frac{{qEd}}{ {m}. {v}_{0}^{2}}$ $[ \text{slope} = \frac{- {qEd}}{ {mv}_{0}^{2}}]$ Now we have to find eq $^{ {n}}$ of straight line whose slope is $\frac{- {qEd}}{ {mv}_{0}^{2}}$ and it pass throughpoint $\to ( {d},- {y}_{0} )$ Because after $x>d$ No electric field $\Rightarrow{F}_{ {\text{net}}}=0, {\vec{v}}= {\text{const}.}$ {y}= {mx}+ {c}, \begin{cases}{m}=\frac{{qEd}}{ {mv}_{0}^{2}} \\ ( {d} & - {y}_{0} ) \} - {y}_{0}= \frac{- {qEd}}{ {mv}_{0}^{2}}\end{cases}{d}+ {c} $\Rightarrow {c}=- {y}_{0}+\frac{{qEd}^{2}}{ {mv}_{0}^{2}}$ Put the value ${y}= \frac{- {qEd}}{{mv}_{0}^{2}} {x}- {y}_{0}+\frac{{qEd}^{2}}{ {mv}_{0}^{2}}$ ${y}_{0}= \frac{1}{2} .\frac{ {qE}}{ {m}} (\frac{{d}}{ {v}_{0}} )^{2}$ $= \frac{1}{2}\frac{{qEd}^{2}}{ {mv}_{0}^{2}}$ $y= \frac{- {qEdx}}{ {mv}_{0}^{2}}- \frac{1}{2} \frac{{qEd}^{2}}{ {mv}_{0}^{2}}+\frac{{qEd}^{2}}{ {mv}_{0}^{2}}$ ${y}= \frac{- {qEd}}{ {mv}_{0}^{2}} {x}+ \frac{1}{2} \frac{{qEd}^{2}}{ {mv}_{0}^{2}}$ $[ {y}=\frac{{qEd}}{ {mV}_{0}^{2}}(\frac{{d}}{2}- {x} )]$

Potential of $-\ {q}$ is same as initial and final point of the path therefore potential due to $4 \ {q}$ will only change and as potential is decreasing the energy will decrease Decrease in potential energy $=\ {q}({V}_{ {i}}- {V}_{ {f}} )$ Decrease in potential energy $= {q} [ \frac{ {k} 4 {q}}{ {d} / 2}-\frac{{k} 4 {q}}{3 {d} / 2} ]= \frac{4 {q}^{2}}{3 \pi \epsilon _{0} {d}}$ Therefore correct answer is 3 .

$E_1={KQ_1}/R_1^2$, $E_2={KQ_2}/R_2^2$Given, $E_1/E_2=R_1/R_2$ $\frac{{KQ_1}/R_1^2}{ {KQ_2}/R_2^2 }=R_1/R_2$ $\Rightarrow Q_1/Q_2=R_1^3/R_2^3$ $V_1/V_2=\frac{KQ_1/R_1}{KQ_2/R_2}=R_1^2/R_2^2$

Thin infinite uniformly charged planes produces uniform electric field therefore option 2 and option 3 are obviously wrong. And as positive charge density is bigger in magnitude so its field along Y direction will be bigger than field of negative charge in $\ {X}$direction and this is evident in option 1 so it is correct.

Let the charges on inner and outer spheres are $\ {Q}_{1}$ and $\ {Q}_{2}$since charge density '\sigma ' is same for bothspheres, so $\sigma =\frac{{Q}_{1}}{4 \pi {r}^{2}}=\frac{{Q}_{2}}{4 \pi {R}^{2}}$ $\Rightarrow \frac{ {Q}_{1}}{ {Q}_{2}}=\frac{{r}^{2}}{ {R}^{2}}$ ${Q}_{1}+ {Q}_{2}= {Q} \Rightarrow \frac{{Q}_{2} {r}^{2}}{ {R}^{2}}+ {Q}_{2}= {Q}$ $\Rightarrow {Q}_{2}=\frac{{QR}^{2}}{({r}^{2}+ {R}^{2} )}$ ${Q}_{1}=\frac{{r}^{2}}{ {R}^{2}} .\frac{{QR}^{2}}{({R}^{2}+ {r}^{2} )}$ $=\frac{{Qr}^{2}}{({R}^{2}+{r}^{2} )}$ Potential at centre 'O' $=\frac{{kQ}_{1}}{ {r}}+\frac{{kQ}_{2}}{ {R}}$ $= {k} [\frac{{Qr}^{2}}{ {r}({R}^{2}+ {r}^{2} )}+\frac{{QR}^{2}}{ {R}({R}^{2}+ {r}^{2} )} ]$ $=\frac{{kQ}( {r}+ {R})}{({R}^{2}+ {r}^{2} )}$ $= \frac{1}{4 \pi \in _{0}} \frac{( {R}+ {r})}{({R}^{2}+ {r}^{2} )} {Q}$

Electric field one to A and D together will be along -ve x-axis, since, the vertical components get cancelled. Similarly, due B and C, in the +ve X- axis Net electric field due to A and D $=2 \times \frac{k q}{D^{2}+4 d^{2}} \times {D(- \hat{i})}/{\sqrt{D^{2}+4 d^{2}}} \ {A} {t} \ {P}$ Electric field one to B and C at P $=2 \\times \frac{k q}{D^{2}+d^{2}} \\times {D}/{\\sqrt{D^{2}+d^{2}}}( \hat{i})$ $\therefore$ Net electric field at $P=\ \vec{E}=2 k q D (\ \frac{1}{\ (D^{2}+d^{2}\ )^{3 / 2}}-\ \frac{1}{\ (D^{2}+d^{2}\ )^{3 / 2}}\ ) \ \hat{i}$ Since, d<<< D, \ \vec{E} \approx 2 K q D\ (\ {1}/{D^{3}\ (1+\ \frac{d^{2}}{D^{2}}\ )^{3 / 2}}-\ {1}/{D^{2}\ (1+\ \frac{4 d^{2}}{D^{2}} )^{3 / 2}}\ ) \ {i} ^{\^} $\approx \frac{2 K q D}{D^{3}}\ (1-\ \frac{3}{2} \frac{d^{2}}{D^{2}}-1+\ \frac{3}{2} \times \frac{4 d^{2}}{D^{2}}\ ) \ \hat{i}$ $\ \therefore E \propto \frac{1}{D^{4}}$

|$P_1$| = q(d) |$P_2$| = qd |Resultant| = 2 P cos30º 2qd $(\sqrt3/2)$ = $\sqrt3$ qd

Electric field of equitorial plane of dipole = - ${K\vec{P}}/r^3$ ∴ At P , F = - ${K\vec{P}}/r^3$ At $P^1 , F^1$ = - ${K\vec{P}Q}/(r/3)^3$ = 27F

At any instant 't' Total energy of charge distribution is constant i.e. $1/2mV^2$ + $\frac{KQ^2}{2R}$ = 0 + $\frac{KQ^2}{2R}$ ∴ $1/2 mV^2$ = $\frac{KQ^2}{2R_0}$ - $\frac{KQ^2}{2R}$ ∴ V = $\sqrt{2/m {KQ^2}/2 . (1/R_0-1/R)}$ ∴ V = $\sqrt{ {KQ^2}/m . (1/R_0-1/R)}$ = C $\sqrt{1/R_0-1/R}$ Also the slope of v-s curve will go ondecreasing ∴ Graph is correctly shown by option (1)

Potential difference doesn't change.

$\tan \theta = \frac{E q}{m g}=0.5\Rightarrow \theta =\tan ^{-1}(0.5)$

$\Delta V=-∫\vec{E} .d x \vec{i}$ $V_{1}-V_{2}=-∫_{-5} ^{1}(A x+B) d x$ = \[\frac{A x^{2}}{2}+B x]^{-5} _{1}$$= (10(5)^{2}-50 )-(10+10)$$\Rightarrow 200-20=180 V$

Q = ∫ ρdv = $∫^{R}_{0} A/r^2 e^{-2r/a}$ $(4\pi r^2dr)$ = 4πA $∫^{R}_{0} e^{-2r/a}$ dr = 4πA $(e^\frac{-2r/a}{-2/a})^R_0$ = 4πA $(-a/2)$ ($e^{-2R/a}$ - 1) Q = 2πaA (1 - $e^{-2R/a}$) R = $a/2$ log $(1/{1-Q/{2\pi aA}})$

$\rho (r)=K r \;Q_{ {total}}=2 Q=∫_{0}^{R} \rho (r)\ (4 \pi r^{2} d r\ )$ $2 Q=4 \pi k \int _{0}^{R} r^{3} d r$ $\;2 Q=4 \pi k\ (\ \frac{R^{4}}{4}\ )$ $Q= \frac{4 k R^{2}}{2} \Rightarrow K=\ \frac{2 Q}{\pi R^{4}}$ Force between charges $-\ {Q}$ and $\ {Q} F=\ \frac{1}{4 \pi ∊_{0}} \frac{Q^{2}}{(2 a)^{2}}=\ \frac{1}{4 \pi \∊_{0}} \ \frac{Q^{2}}{4 a^{2}}$ $E\int d s=\ \frac{q_{i n}}{\∊_{0}} E\ (4 \pi a^{2}\ )=∫_{0}^{a} {\rho d v}{\∊_{0}}$ $E=\ \frac{Q a^{2}}{2 \pi \∊_{0} R^{4}} F=Q E \frac{Q^{2}}{16 \pi ∊_{0} a^{2}}=\ \frac{Q^{2} a^{2}}{2 \pi \∊_{0} R^{n}}$ $a=8^{-1 / 4} R$

U = - $\vec{P}.\vec{E}$ = - PE cos θ = - $(10^{-29})$ $(10^3)$ cos 45° = - 0.707 × $10^{-26}$ J = - 7 × $10^{-27}$ J

The potential energy of +q, -q system is $U_{1}=\ \frac{-k q^{2}}{d}$ The system of +q, -q act as an electric dipole as d<< D Hence, Potential energy between Q and dipole is $U_{2}=\ \frac{-k p}{r^{2}} Q$Hence, $U=U_{1}+U_{2}=\ \frac{+1}{4 \pi \epsilon _{0}}\ [\ \frac{-q^{2}}{d}-\ \frac{d \theta d}{D^{2}}\ ]$