Electrostatics Practice Questions

Conceptual

Electrons moves in curved path because there velocity ${ {\text{u}}}^{\to }$ may make any angle $\theta$ with acceleration $\vec{a}$ between time interval of two successive collisions.Also electron moves from lower potential to higher potential.

\;{ \vec{\text{E}}=\;{ {\text{A}}}/{ {\text{x}}^2} \;{ {\text{i}}}^{\^}+\;{ {\text{B}}}/{ {\text{y}}^3} \;{ {\text{j}}}^{\^} \;{[\;{ {\text{A}}}/{ {\text{x}}^2}]\ = {\text{NC}}^{-1} \Rightarrow [ {\text{A}}]= {\text{Nm}}^2 {\text{C}}^{-1}}$$;{[;{ {\text{B}}}/{ {\text{y}}^3}]$ = {\text{NC}}^{-1} \Rightarrow [ {\text{B}}]= {\text{Nm}}^3 {\text{C}}^{-1}}$

In uniform circular motion$\;F_c=m a_c$ $\;\;\text{ (q) }\;(E)=\;{ {\ mv}^2}/{ {\text{r}}}$(e) $(\;{2 {\text{k}} \lambda }/{ {\text{r}}}) =\;{ {\ mv}^2}/{ {\text{r}}}$ $\; {\text{v}}^2=\;{( {\text{e}})(2 {\text{k}} \lambda )}/{ {\text{m}}}=\;\frac{ (1.6 \times 10^{-19}) \times 2 \times (9 \times 10^9) \times (2 \times 10^{-8}) }{9 \times 10^{-31}}$ $\;=1.6 \times 4 \times 10^{13}$ $\; {\text{V}}^2=16 \times 4 \times 10^{12} \Rightarrow {\text{v}}=8 \times 10^6 {\text{m}} / {\ s}$Ans. 8

$\;\;\frac{Q_1^{'}}{4 \pi \epsilon _0 R}=\;\frac{Q_2^{'}}{4 \pi \epsilon _0(2 R)}$ $\therefore \;\; Q_2^{'}=2 Q_1^{'}$ $\;Q_1^{'}+Q_2^{'}=Q_1+Q_2$ $\therefore \;\;\frac{Q_2^{'}}{2}+Q_2^{'}=20 \pi R^2 \sigma$ $\;\;\frac{3}{2} Q_2^{'}=20 \pi R^2 \sigma$ $\therefore \;\;\frac{Q_2^{'}}{4 \pi (2 R)^2}=\;\frac{2}{3} \cdot \;\frac{20 \pi R^2 \sigma }{16 \pi R^2}$ $\therefore \;\;\frac{\sigma ^{'}}{\sigma }=\;\frac{5}{6}$

{ \vec{\text{E}}=2 {\text{x}}^2 \;{ {\text{i}}}^{\^}-4 {\text{y}} \;{ {\text{j}}}^{\^}+6 \;{ {\text{k}}}^{\^} $\;\varphi _{\;\text{net }\;}=-8 \times 3+2 \times 6=-12$ $\;-12=\;{ {\text{q}}}/{E_0}$ $\;| {\text{q}}|=12 E_0$

$E=\;\frac{K P}{r^3} \propto \;\frac{1}{r^3}$

(i) An electric dipole is enclosed in a closed question surface the total flux through the enclosed surface is zero.(ii) net charge inside the enclosed surface is zero.

\;\text{ Flux }\;\;={ \vec{\text{E}} \cdot { \vec{\text{A}} $\;=4000(0 \cdot 2)^2 \;{ {\text{V}}}/{ {\text{m}}} \cdot (0 \cdot 2)^2 {\text{m}}^2$ $\;=4000 \times 16 \times 10^{-4} {\text{Vm}}$ $\;=640 {\text{Vcm}}$Ans. 640

\;{ \vec{\text{E}}= {\text{E}}_0 \times \;{ {\text{i}}}^{\^} $\;\varphi _{\;\text{net }\;}=\varphi _{ {\text{ABCD}}}= {\text{E}}_0 {\text{a}} \cdot {\text{a}}^2$ $\;\;{ {\text{q}}_{\;\text{en }\;}}/{E_0}= {\text{E}}_0 {\text{a}}^3$ $\; {\text{q}}_{ {\text{en}}}= {\text{E}}_0 E_0 {\text{a}}^3$ $\;=4 \times 10^4 \times 9 \times 10^{-12} \times 8 \times 10^{-6}$ $\;=288 \times 10^{-14} {\text{C}}$ $\; {\text{Q}}=288$Ans. 288

Point charge on equilibrium is at rest.$\;\;\text{ So }\; \;\; {\text{F}}_{ {\text{e}}}= {\text{mg}} sin \;\theta$ $\;\;{ {\text{kq}}_0 \cdot {\text{q}}_0}/{ {\text{r}}^2}= {\text{mg}} sin \;30^{\circ}$ $\;\;{ {\text{kq}}_0^2}/{ (\;{ {\text{h}}}/{ sin \;30^{\circ}}) ^2}=\;{ {\text{mg}}}/{2}$ $\;\;{9 \times 10^9 \times (2 \times 10^{-6}) ^2}/{4 {\text{h}}^2}=\;\frac{20 \times 10^{-3} \times 10}{2}$ $\;\;{9 \times 4 \times 10^9 \times 10^{-12}}/{4 {\text{h}}^2}=10^{-1}$ $\; {\text{h}}^2=9 \times 10^{-2} \;\; \;\text{ Ans. }\;$ $\; {\text{h}}=0.3 {\text{m}}=300 \times 10^{-3} {\text{m}}$ $\; {\text{x}}=300 \;\; \;\text{ Ans. }\;$

Assuming RHS to be $\hat{n}$ $\;\vec{E}_{ {\text{I}}}=\;\frac{\sigma }{2 E_0}(-\hat{n})+\;\frac{\sigma }{2 E_0}(-\hat{n})=-\;\frac{\sigma }{E_0}\hat{n}$ $\;\vec{E}_{I I}=0$ $\;\vec{E}_{I I I}=\;\frac{\sigma }{2 E_0}(\hat{n})+\;\frac{\sigma }{2 E_0}(\hat{n})=\;\frac{\sigma }{E_0}(\hat{n})$

$\varphi =\;\frac{Q / \epsilon _0}{6}$Flux passing through shaded face $=\;{ {\text{q}}}/{6 \epsilon _0}$

$E_p=\;\frac{K \times 10}{2^2}-\;\frac{K \times 40}{ (x_0-2) ^2}=0$ $\;\frac{1}{2}=\;\frac{2}{x_0-2}$ $x_0-2=4$ ${\text{x}}_0=6 {\text{cm}}$

Divide $q=10 \mu$ cinto parts $(x)$ and $(q-x)$ ${\text{F}}=\;{( {\text{K}})( {\text{x}})( {\text{q}}- {\text{x}})}/{ {\text{r}}^2}$For ${\text{F}}$ to be maximum$\;\;\frac{d F}{d x}=0$ $\;x=\;\frac{q}{2}=\;\frac{10 \mu C}{2}=5 \mu C$ $\;q-x=10 \mu C-5 \mu C=5 \mu C$

$\; {\text{W}}_{\;\text{ext }\;}= {\text{U}}_{ {\text{f}}}- {\text{U}}_{ {\text{i}}} \;\; \{ {\text{U}}=-{ {\text{P}}}^{\to } \cdot { {\text{E}}}^{\to }\}$ $\;=- {\text{PE}} \cos \pi -(- {\text{PE}} \cos 0)$ $\;=2 {\text{PE}}$ $\;=2 \times 6 \times 10^{-6} \times 1.5 \times 10^3$ $\;=18 {\text{mJ}}$

$\; {\text{V}}=\;\frac{1}{4 \pi \epsilon _0} \;{ {\text{q}}}/{ {\text{r}}}$ $\;\Rightarrow {\text{r}}=\;\frac{9 \times 10^9 \times 5 \times 10^{-9}}{50}$ $\;\Rightarrow {\text{r}}=\;\frac{9}{10} \times 100 {\text{cm}}$ $\; {\text{r}}=90 {\text{cm}}$

Electric field due to uniformly charged insulating solid sphere

$\;0.5 {\text{e}}=\;\frac{1}{2} {\ mv}_{ {\text{x}}}^2 \Rightarrow {\text{v}}_{ {\text{x}}}=\sqrt{\;{ {\text{e}}}/{ {\text{m}}}}$ $\;\;\text{ Along }\; x {\text{L}}= {\text{v}}_{ {\text{x}}} {\text{t}}=\sqrt{\;{ {\text{e}}}/{ {\text{m}}}} {\text{t}}$ $\;\;\text{ Along y }\; {\text{v}}_{ {\text{y}}}=\;{ {\text{eE}}}/{ {\text{m}}} {\text{t}}$ $\;{\text{dividing}} \;{ {\text{v}}_{ {\text{y}}}}/{ {\text{L}}}= {\text{E}} \sqrt{\;{ {\text{e}}}/{ {\text{m}}}}= {\text{Ev}}_{ {\text{x}}}$ $\;\Rightarrow {\text{Tan}} \theta =\;{ {\text{v}}_{ {\text{y}}}}/{ {\text{v}}_{ {\text{x}}}}= {\text{E}} \times {\text{L}}=10 \times 0.1=1$ $\;\theta =45^{\circ}$ $\;$