Electrostatics Practice Questions

The distance of point $A({\sqrt{2}}, {\sqrt{2}})$ from the origin, $O A= |\vec{r_1}{{}/{}|}$ $=\sqrt{({\sqrt{2}})^{2+}({\sqrt{2}})^2}$ $={\sqrt{4}}=2$ The distance of point $B(2,0)$ from the origin, $O B= |\vec{r_2}{{}/{}|} ={\sqrt{(2)^{2+}(0)^2}}=2$ units. Now, potential at $A, V_A=\;\frac{1}{4 \pi E_0} \cdot \;\frac{Q}{(O A)}$ Potential at $B, V_B=\;\frac{1}{4 \pi E_0} \cdot \;\frac{Q}{(O b)}$ $\therefore$ Potential difference between the points $A$ and $B$ is zero.

As shown in the figure, the resultant electric fields before and after interchanging the charges will have the same magnitude, but opposite directions. Also, the potential will be same in both cases as it is a scalar quantity.

Here, $V(x)=\;\frac{20}{x^{2-}4}$ voltWe know that $E=-\;\frac{d V}{d x}=\;\frac{d}{d x} (\;\frac{20}{x^{2-}4})$ or, $E=+\;\frac{40 x}{ (x^{2-}4) ^2}$At $x=4 \mu m$,$\;E=+\;\frac{40 \times 4}{ (4^{2-}4) ^2}$ $\;=+\;\frac{160}{144}=+\;\frac{10}{9}$ volt${/} \mu {\text{m}} .$Positive sign indicates that $\vec{E}$ is in $+v e x$-direction.

The electric field will be different at the location of the two charges. Therefore the two forces will be unequal. This will result in a force as well as torque.

After connection, $V_1=V_2$ $\;\Rightarrow K \;\frac{Q_1}{r_1}=K \;\frac{Q_2}{r^2}$ $\;\Rightarrow \;\frac{Q_1}{r_1}=\;\frac{Q_2}{r_2}$The ratio of electric fields$\;\;\frac{E_1}{E_2}=\;\frac{K \frac{Q_1}{r_1^2}}{K \;\frac{Q_2}{r_2^2}}=\;\frac{Q_1}{r_1^2} \times \;\frac{r_2^2}{Q_2}$ $\;\Rightarrow \;\frac{E_1}{E_2}=\;\frac{r_1 \times r_2^2}{r_1^2 \times r_2} \Rightarrow \;\frac{E_1}{E_2}=\;\frac{r_2}{r_1}=\;\frac{2}{1}$Since the distance between the spheres is large as com effects may be ignored.

$e V=\;\frac{1}{2} m v^2$ $\Rightarrow v=\sqrt{\;\frac{2 e v}{m}}=\sqrt{\;\frac{2 \times 1.6 \times 10^{-19} \times 20}{9.31 \times 10^{-31}}}$ $=2.65 \times 10^6 {\text{m}} / {\ s}$

$\;\frac{-K 2 q}{(x-L)^2}+\;\frac{K 8 q}{x^2}=0 \Rightarrow \;\frac{1}{(x-L)^2}=\;\frac{4}{x^2}$or, $\;\frac{1}{x-L}=\;\frac{2}{x} \Rightarrow x=2 x-2 L$or, $x=2 L$

$V_A=V_{\;\text{self }\;}+V_{\text{due}}$ to $(2)$ $\Rightarrow V_A=\;\frac{1}{4 \pi \epsilon _0} [\;\frac{q}{R}-\;{q}/{{\sqrt{R^{2+}d^2}}}]$ $V_B=V_{\text{self}}+V_{\text{due}}$ to $(1)$ $\Rightarrow V_B=\;\frac{1}{4 \pi \epsilon _0} [\;\frac{-q}{R}+\;{q}/{{\sqrt{R^{2+}d^2}}}]$ $\Delta V=V_A-V_B$ $=\;\frac{1}{4 \pi \epsilon _0} [\;\frac{q}{R}+\;\frac{q}{R}-\;{q}/{{\sqrt{R^{2+}d^2}}}-\;{q}/{{\sqrt{R^{2+}d^2}}}]$ $=\;\frac{1}{2 \pi \epsilon _0} [\;\frac{q}{R}-\;{q}/{{\sqrt{R^{2+}d^2}}}]$

$\;T sin \;\theta =\;\frac{\sigma }{\epsilon _0 K} \cdot q$....(i)$\;T \cos \theta =m g$...(ii)Dividing $(i)$ by $(i i)$,$\;\tan \theta =\;\frac{\sigma q}{\epsilon _0 K \cdot m g}$ $\; \therefore \sigma \propto \tan \theta$

$F \propto \;\frac{Q_A Q_C}{x^2}$ $x$ is distance between the spheres. After first operation charge on $B$ is halved i.e $\;\frac{Q}{2}$. and charge on third sphere becomes $\;\frac{Q}{2}$. Now it is touched to $C$, charge then equally distributes them selves to make potential same, hence charge on $C$ becomes $(Q+\;\frac{Q}{2}) \;\frac{1}{2}=\;\frac{3 Q}{4}$ $\therefore F_{n e w} \propto \;\frac{Q_C Q_B}{x^2}$ $=\;\frac{ (\frac{3 Q}{4}) (\frac{Q}{2}) }{x^2}$ $=\;\frac{3}{8} \;\frac{Q^2}{x^2}$or, $F_{\;\text{new }\;}=\;\frac{3}{8} F$

At equilibrium, electric force on drop balances weight of drop.$\;q E=m g \Rightarrow q$ $\;=\;\frac{m g}{E}=\;\frac{9.9 \times 10^{-15} \times 10}{3 \times 10^4}$ $\;=3.3 \times 10^{-18} C$

Net field at A should be zero${\sqrt{2}} E_1+E_2=E_3$ $\; \therefore \;{k Q \times {\sqrt{2}}}/{a^2}+\;{k Q}/{({\sqrt{2}} a)}=\;{k q}/{ (\;{a}/{{\sqrt{2}}}) ^2}$ $\;\Rightarrow \;{Q {\sqrt{2}}}/{1}+\;\frac{Q}{2}=2 q$ $\;\Rightarrow q=\;\frac{Q}{4}(2 {\sqrt{2}}+1)$

$\;\frac{1}{2} m v^2=\;\frac{k Q q}{r}$ $\Rightarrow \;\frac{1}{2} m(2 v)^2=\;\frac{k q Q}{r^{'}}$ $\Rightarrow r^{'}=\;\frac{r}{4}$

The flux entering an enclosed surface is taken as negative and the flux leaving the surface is taken as positive, by convention. Therefore the net flux leaving the enclosed surface $=\varphi _2-\varphi _1$ $\therefore$ the change enclosed in the surface by Gauss's law is $q=\epsilon _0 (\varphi _2-\varphi _1)$

Force on charge $q_1$ due to $q_2$ is $F_{12}=k \;\frac{q_1 q_2}{b^2}$Force on charge $q_1$ due to $q_3$ is $F_{13}=k \;\frac{q_1 q_3}{a^2}$The $X$-component of the force $(F_x)$ on$q_1$ is $F_{12}+F_{13} sin \;\theta$ $\therefore F_x=k \;\frac{q_1 q_2}{b^2}+k \;\frac{q_1 q_2}{a^2} sin \;\theta$ $\therefore F_x \propto \;\frac{q_2}{b^2}+\;\frac{q_3}{a^2} sin \;\theta$

Electric potential due to charge $Q$ placed at the center of spherical shell at point $P$ is$V_1=\;\frac{1}{4 \pi \epsilon _0} \;\frac{Q}{R {/} 2}=\;\frac{1}{4 \pi \epsilon _0} \;\frac{2 Q}{R}$Electric potential due to charge $q$ on the surface of the spherical shell at any point inside the shell is$V_2=\;\frac{1}{4 \pi \epsilon _0} \;\frac{q}{R}$

Both the charges are identical and placed symmetrically about $A B C D$. The flux crossing $A B C D$ due to each charge is $\;\frac{1}{6}[\;\frac{q}{E_0}]$ but in opposite directions. Therefore the resultant is zero.

For equilibrium of charge $Q$ $K \;\frac{Q \times Q}{(2 x)^2}+K \;\frac{Q q}{x^2}=0 \Rightarrow q=-\;\frac{Q}{4}$

We know that $\;\frac{W_{A B}}{q}=V_B-V_A$ $\therefore V_B-V_A=\;{2 {\text{J}}}/{20 C}=0.11 {\text{J}} {/} C=0.1 {\text{V}}$