Electrostatics Practice Questions

\;{ {\text{E}}}^{\to }=6 { {\text{i}}^{\^}}+5 { {\text{j}}^{\^}}+3 { {\text{k}}^{\^}} \;{ {\text{A}}}^{\to }=30 { {\text{i}}^{\^}} $\;\phi ={ {\text{E}}}^{\to } \cdot { {\text{A}}}^{\to }$ \;\phi =(6 { {\text{i}}^{\^}}+5 { {\text{j}}^{\^}}+3 { {\text{k}}^{\^}}) \cdot (30 { {\text{i}}^{\^}}) $\;\phi =6 \times 30=180$

{ {\text{E}}}^{\to }_{ {\text{p}}}\;= (\;\frac{\sigma }{2 \epsilon _0}+\;\frac{2 \sigma }{2 \epsilon _0}+\;\frac{\sigma }{2 \epsilon _0}) (-{ {\text{i}}^{\^}}) \;=-\;\frac{2 \sigma }{\epsilon _0} { {\text{i}}^{\^}}

$\;\vec{\tau }={ {\text{p}}}^{\to } \times { {\text{E}}}^{\to }$ $\;{ {\text{p}}}^{\to }= {\text{q}} \vec{ℓ}$ $\;{ {\text{E}}}^{\to }=0.2 \;{ {\text{V}}}/{ {\text{cm}}}=20 \;{ {\text{V}}}/{ {\text{m}}}$ \;{ {\text{p}}}^{\to }=4 \times ({ {\text{i}}^{\^}}-{ {\text{j}}^{\^}}+{ {\text{k}}^{\^}}) \;=(4 { {\text{i}}^{\^}}-4 { {\text{j}}^{\^}}+4 { {\text{k}}^{\^}}) \mu {\text{C}}- {\text{m}} \;\vec{\tau }=(4 { {\text{i}}^{\^}}-4 { {\text{j}}^{\^}}+4 { {\text{k}}^{\^}}) \times (20 { {\text{i}}^{\^}}) \times 10^{-6} {\text{Nm}} \;=(8 { {\text{k}}^{\^}}+8 { {\text{j}}^{\^}}) \times 10^{-5}=8 {\sqrt{2}} \times 10^{-5} $\;\alpha =2$

$\;E_P=\;\frac{2 K P}{r^3}=E$ $\;E_R=\;\frac{K P}{(2 r)^3}=\;\frac{E}{16}$ $\;x=16$

\;{ {\text{E}}}^{\to }=2 \times { {\text{i}}^{\^}} $\;\varphi ={ {\text{E}}}^{\to } \cdot { {\text{A}}}^{\to }$ $\;\varphi _{\;\text{\in }\;}=-4 \times 4=-16 {\text{Nm}}^2 / {\text{c}}$ $\;\varphi _{\;\text{out }\;}=8 \times 4=32 {\text{Nm}}^2 / {\text{c}}$ $\; {\text{d}}_{\;\text{net }\;}=\varphi _{\;\text{\in }\;+} \varphi _{\;\text{out }\;}=-16+32=16 {\text{Nm}}^2 / {\text{c}}$

Static charge is developed due to air friction. This can result in combustion. So, metallic chains is used to discharge excess charge.

5Q charge is inside the spherical region flux through sphere $=\;{5 {\text{Q}}}/{\epsilon _0}$

$\; (\vec{E}_{\;\text{net }\;}) _P=0$ $\;\;\frac{k q}{x^2}=\;\frac{k \cdot 3 q}{(r-x)^2}$ $\;(r-x)^2=3 x^2$ $\;r-x={\sqrt{3}} x$ $\;x=\;{r}/{{\sqrt{3}}+1}$

$\;\;{2 {\text{k}} \lambda {\text{q}}}/{ {\text{r}}}= {\text{m}} \omega ^2 {\text{r}}$ $\;\omega ^2=\;{2 {\text{k}} \lambda {\text{q}}}/{ {\text{mr}}^2}$ $\; (\;{2 \pi }/{ {\text{T}}}) ^2=\;{2 {\text{k}} \lambda {\text{q}}}/{ {\text{mr}}^2}$ $\; {\text{T}}=2 \pi {\text{r}} \sqrt{\;{ {\text{m}}}/{2 {\text{k}} \lambda {\text{q}}}}$

$\; {\text{u}}=1 {\text{m}} / {\ s} ; {\text{a}}=-\;{\sigma {\text{e}}}/{2 \epsilon _0 {\text{m}}}$ $\; {\text{t}}=1 {\ s}$ $\; {\text{S}}=-1 {\text{m}}$ $\;\;\text{ Using }\; {\text{S}}= {\text{ut}}+\;\frac{1}{2} {\text{at}}^2$ $\;-1=1 \times 1-\;\frac{1}{2} \times \;{\sigma {\text{e}}}/{2 \epsilon _0 {\text{m}}} \times (1)^2$ $\; \therefore \sigma =8 \;{\epsilon _0 {\text{m}}}/{ {\text{e}}}$ $\; \therefore \alpha =8$

In air $\tan \;\frac{\theta }{2}=\;{ {\text{F}}}/{ {\text{mg}}}=\;{ {\text{q}}^2}/{4 \pi \epsilon _0 {\text{r}}^2 {\text{mg}}}$ In water $\tan \;\frac{\theta }{2}=\;{ {\text{F}}^{'}}/{ {\text{mg}}^{'}}=\;{ {\text{q}}^2}/{4 \pi \epsilon _0 \epsilon _{ {\text{r}}} {\text{r}}^2 {\text{mg}}_{\;\text{eff }\;}}$ Equate both equations $\;\epsilon _0 g=\epsilon _0 \epsilon _{ {\text{r}}} {\text{g}} [1-\;\frac{1}{1.5}]$ $\;\epsilon _{ {\text{r}}}=3$

$\;\phi _{\;\text{smaller cube }\;}=\;{2 {\text{Q}}}/{\in _0}$ $\;\phi _{\;\text{bigger cube }\;}=\;{5 {\text{Q}}}/{\in _0}$ $\;\;{\phi _{\;\text{smaller cube }\;}}/{\phi _{\;\text{bigger cube }\;}}=\;\frac{2}{5}$

$\;\;\text{ Potential }\;=\;\frac{k Q}{R}=\;\frac{k \cdot Z e}{R}$ $\;=\;\frac{9 \times 10^9 \times 50 \times 1.6 \times 10^{-19}}{9 \times 10^{-13} \times 10^{-2}}$ $\;=8 \times 10^6 {\text{V}}$

$V=\;\frac{Σ K Q_i}{r_i}$Here, $Q_i$ and $r_i$ are positive.$\therefore V>0$The correct statement is:(A) Net potential of the system cannot be zero at a point but net electric field can be zero at that point.Explanation:In a group of positive charges, the net potential at a point is the sum of the potentials due to each individual charge. The potential due to a point charge is given by the Coulomb's law, which is non-zero except at the location of the charge itself. Therefore, the net potential due to a group of positive charges can never be zero at a point.On the other hand, the net electric field at a point is the vector sum of the electric fields due to each individual charge. If the charges are arranged in such a way that their electric fields cancel out at a particular point, then the net electric field at that point can be zero, even though the charges are present. This can happen, for example, in a symmetrical arrangement of charges.

$\;\;{ {\text{q}}_0}/{ {\text{x}}^2}=\;{4 {\text{q}}_0}/{( {\text{x}}+12)^2}$ $\; {\text{x}}+12=2 {\text{x}}$ $\; {\text{x}}=12$Distance from origin $=x+12=24 {\text{cm}}$.

$\; {\text{E}}_{ {\text{p}}}=\;{2 {\text{K}} \lambda }/{ {\text{r}}}-\;\frac{\sigma }{2 \epsilon _0}$ $\; {\text{E}}_{ {\text{p}}}=\;\frac{\sigma }{2 \epsilon _0}-\;\frac{\lambda }{2 \pi \epsilon _0 (\;\frac{3}{\pi }) }$ $\; {\text{E}}_{ {\text{p}}}=\;\frac{2 \sigma }{2 \epsilon _0}-\;\frac{2 \sigma }{6 \epsilon _0}=\;\frac{\sigma }{6 \epsilon _0}$ $\;\;\text{ Similarly, }\; {\text{E}}_{ {\text{Q}}}=\;\frac{\sigma }{2 \epsilon _0}-\;\frac{2 \sigma }{2 \pi \epsilon _0 (\;\frac{4}{\pi }) }=\;\frac{\sigma }{4 \epsilon _0}$ $\;\;{ {\text{E}}_{ {\text{p}}}}/{ {\text{E}}_{ {\text{Q}}}}=\;\frac{4}{6}$ $\; {\text{a}}=6$

\;\omega _{ {\text{E}}}={q \vec{\text{E}} \cdot { \vec{\text{S}} \;=2 \times 10^{-2}[30 \;{ {\text{i}}}^{\^} \cdot (-\;{ {\text{i}}}^{\^})] $\;=2 \times 10^{-2}(-30)$ $\;=-60 \times 10^{-2}$ $\;=-\;\frac{60}{100}=-0.6 {\text{J}}$ $\;=-600 {\text{mJ}}$

$\; {\text{a}}=\;{ {\text{F}}}/{ {\text{m}}}=\;{ {\text{qE}}}/{ {\text{m}}}= (2 \times 10^{11}) (1.8 \times 10^3)$ $\;=3.6 \times 10^{14} {\text{m}} / {\ s}^2$ $\;\;\text{ Time to cross plates }\;=\;{ {\text{d}}}/{ {\text{v}}}$ $\; {\text{t}}=\;\frac{0.10}{3 \times 10^7}$ $\; {\text{y}}=\;\frac{1}{2} {\text{at}}^2=\;\frac{1}{2} (3.6 \times 10^{14}) (\;\frac{0.01}{9 \times 10^{14}})$ $\;=0.2 \times 0.01$ $\;=0.002 {\text{m}}$ $\;=2 {\text{mm}}$