Electrostatics Practice Questions

According Gauss's law, we can write,E\times 4\pi r^2=1/\epsilon _0\[Q+ ∫^r_a(A/r)(4πr^2dr)]$=1/\epsilon _0[Q+2\pi A(r^{2-}a^2)]$For E to be independent of r,$Q-2\pi Aa^2=0\Rightarrow a=Q/{2\pi a^2}$

$\frac{\Delta V}{\Delta r}\to$constant\Rightarrow unifor E.field$(E)(4\pi r^2)=1/\epsilon _0∫\rho dV$ $(E)(4\pi r^2)=1/\epsilon _0∫^{r}_{0}\rho 4\pi r^2dr$ $(E)(4\pi r^2)=1/\epsilon _{0}4\pi ∫^{r}_{0}\rho r^2dr$after integral on RHSWe must obtain $r^2$ $\Rightarrow \rho \propto 1/r$

It originates from +Ve charge and terminates at - Ve charge. It can not form close loop.

The potential at the centre $=kQ/{4/3\pi R^3}∫^R_0{4\pi r^3dr}/r=3/2{kQ}/R=3/2V_0;k=1/{4\pi \epsilon _0}$ $\therefore R_1=0$Potential at surface,$V_0={KQ}/R$Potential at $R_2={5V_0}/4\Rightarrow R_2=R/\sqrt2$Potential at $R_3,{kQ}/R_3=3/4{KQ}/R\Rightarrow R_3={4R}/3$Similarly at $R_4,{kQ}/R_4=\frac{kQ}{4R}\Rightarrow R_4=4R$∴ Both options (2) and (3) are correct.

Potential difference between any two points in an electric field is given by, $\;d V=-\vec{E} \cdot \vec{d x}$ $\;∫_{V_{O}}^{V_{A}} d V=-∫_{0}^{2} 30 x^{2} d x$ \;V_{A}-V_{O}=\[10 x^{3}]_{0}^{2}=-80 J / C$

$F_{\text{net}}=2Fcos\theta$ $=2{\text{k\cdot q\cdot q/2}}/(\sqrt{a^{2+}y^2})^2\cdot y/(\sqrt{a^{2+}y^2})$ ={kq^2y}/a^3(ya)

$V=∫^{x=2L}_{x=L}k/x(Q/L)dx=\frac{Qln2}{4\pi \epsilon _0L}$

The electric field inside a uniformly charged sphere is$=\;\frac{\rho . r}{3 E_0}$The electric potential inside a uniformly charged sphere$=\;\frac{\rho R^2}{6 \in _0} [3-\;\frac{r^2}{R^2}]$ $\therefore$ Potential difference between center and surface$\;=\;\frac{\rho R^2}{6 E_0}[3-2]=\;\frac{\rho R^2}{6 E_0}$ $\;\Delta U=\;\frac{q \rho R^2}{6 E_0}$

$E_{i n} \propto r$ $E_{\;\text{out }\;} \propto \;\frac{1}{r^2}$

Electric field$E=\;\frac{d \varphi }{d r}=-2 a r$By Gauss's theorem$E=\;\frac{1}{4 \theta \epsilon _0 r^2}$From $(i)$ and $(i i)$,$\;q=-8 \pi \epsilon _0 a r^3$ $\;\Rightarrow d q=-24 \pi \epsilon _0 a r^2 d r$Charge density, $\rho =\;\frac{d q}{4 \pi r^2 d r}=-6 \epsilon _0 a$

At any instant$\;T \cos \theta =m g \;\; . . .(i)$ $\;T sin \;\theta =F_e \;\; . . .(i i)$ $\;\Rightarrow \;\frac{ sin \;\theta }{\cos \theta }=\;\frac{F_e}{m g} \Rightarrow F_e=m g \tan \theta$ $\;\Rightarrow \;\frac{k q^2}{x^2}=m g \tan \theta \Rightarrow q^2 \propto x^2 \tan \theta$ $\; sin \;\theta =\;\frac{x}{2 l}$ $\;\;\text{ For }\; {\ small} \theta , sin \;\theta \approx \tan \theta$ $\; \therefore q^2 \propto x^3$ $\Rightarrow q \;\frac{d q}{d t} \propto x^2 \;\frac{d x}{d t}$ $\therefore \;\frac{d q}{d t}=$ const.$\therefore q \propto x^2 \cdot v \Rightarrow x^{3 / 2} \alpha x^2 \cdot v \;\; [.$ as $q^2 \propto x^3]$ $\Rightarrow v \propto x^{-1 / 2}$

Let us consider a spherical shell of radius $x$ and thickness $d x$.Charge on this shell $d q=\rho \cdot 4 \pi ^2 \text{dx}=\rho _0 (\;\frac{5}{4}-\;\frac{x}{R}) \cdot 4 \pi x^2 \text{dx}$ $\therefore$ Total charge in the spherical region from center to $r(r< R)$ is $q=∫ d q=4 \pi \rho _0 ∫_{0}^{r} (\;\frac{5}{4}-\;\frac{x}{R}) x^2 \text{dx}$ $=4 \pi \rho _0 [\;\frac{5}{4} \cdot \;\frac{r^3}{3}-\;\frac{1}{R} \cdot \;\frac{r^4}{4}]$ $=\pi \rho _0 r^3 (\;\frac{5}{3}-\;\frac{r}{R})$ $\therefore$ Electric field at $r, E=\;\frac{1}{4 \pi E_0} \cdot \;\frac{q}{r^2}$ $=\;\frac{1}{4 \pi E_0} \cdot \;\frac{\pi \rho _0 r^3}{r^2} (\;\frac{5}{3}-\;\frac{r}{R})$ $=\;\frac{\rho _0 r}{4 E_0} (\;\frac{5}{3}-\;\frac{r}{R})$

Let us consider a differential element $d l$. charge on this element. $d q= (\;\frac{q}{\pi r}) d l$ $=\;\frac{q}{\pi r}(r d \theta ) \;\; ($ as $d l=r d \theta )$ $= (\;\frac{q}{\pi }) d \theta$ Electric field at $O$ due to $d q$ is$\;d E=\;\frac{1}{4 \pi E_0} \cdot \;\frac{d q}{r^2}$ $\;=\;\frac{1}{4 \pi E_0} \cdot \;\frac{q}{\pi r^2} d \theta$ The component $d E \cos \theta$ will be counter balanced by another element on left portion.Hence resultant field at $O$ is the resultant of the component $d E \sin \;\theta$ only. $\; \therefore E=∫ d E sin \;\theta =∫_{0}^{\pi } \;\frac{q}{4 \pi ^2 r^2 E_0} \sin \;\theta d \theta$ $\;=\;\frac{q}{4 \pi ^2 r^2 \in _0}[-\cos \theta ]_0^\pi$ $\;=\;\frac{q}{4 \pi ^2 r^2 \in _0}(+1+1)$ $\;=\;\frac{q}{2 \pi ^2 r^2 \in _0}$The directions of $E$ is towards negative $y$-axis.$\therefore \vec{E}=-\;\frac{q}{2 \pi ^2 r^2 E_0} {j}^{∧}$

$\;F_e=T sin \;15^{\circ}$ $\;m g=T \cos 15^{\circ}$ $\;\Rightarrow \tan 15^{\circ}=\;\frac{F_e}{m g}$......(i)In liquid, $F_e{ }^{'}=T^{'} sin \;15^{\circ}$.....(ii)$m g=F_B+T^{'} \cos 15^{\circ}$ $F_B^{'}=V(d-\rho ) g=V(1.6-0.8) g=0.8 V g$ $=0.8 \;\frac{m}{d} g=\;\frac{0.8 m g}{1.6}=\;\frac{m g}{2}$ $\therefore m g=\;\frac{m g}{2}+T^{'} \cos 15^{\circ}$ $\Rightarrow \;\frac{m g}{2}=T^{'} \cos 15^{\circ}$From $(A)$ and $(B), \tan 15^{\circ}=\;\frac{2 F_e^{'}}{m g}$.......(2)From $(1)$ and $(2)$ $\;\frac{F_e}{m g}=\;\frac{2 F_e^{'}}{m g} \Rightarrow F_e=2 F_e^{'} \Rightarrow F_e^{'}=\;\frac{F_e}{2}$

Let $F$ be the force between $Q$ and $Q$. The force between $q$ and $Q$ should be attractive for net force on $Q$ to be zero. Let $F^{'}$ be the force between $Q$ and $q$. For equilibrium$\;{\sqrt{2}} F^{'}=-F$ $\;{\sqrt{2}} \times k \;\frac{Q q}{ℓ^2}=-k \;{Q^2}/{({\sqrt{2}} ℓ)^2}$ $\;\Rightarrow \;\frac{Q}{q}=-2 {\sqrt{2}}$

Statement 1 is true.Statement 2 is true and is the correct explanation of $(1)$

Let us consider a spherical shell of thickness $d x$ and radius $x$. The volume of this spherical shell $=4 \pi r^2 d r$. The charge enclosed within shell $=\;\frac{Q_r}{\pi R^4} [4 \pi r^2 d r]$ The charge enclosed in a sphere of radius $r_1$ is $\;\frac{4 Q}{R^4} ∫_{0}^{r_1} r^3 d r$ =\;\frac{4 Q}{R^4} \[;\frac{r^4}{4}]$ _0^{r_1}$$=;\frac{Q}{R^4} r_1^4$$\therefore $The electric field at point$p$inside the sphere at a distance$r_1$from the center of the sphere is$;E=;\frac{1}{4 \pi E_0} ;\frac{[\frac{Q}{R^4} r_1^4]$ }{r_1^2}$$;=;\frac{1}{4 π E_0} ;\frac{Q}{R^4} r_1^2$

$\;\frac{W_{P Q}}{q}= (V_Q-V_P)$ $\Rightarrow W_{P Q}=q (V_Q-V_P)$ $= (-100 \times 1.6 \times 10^{-19}) (-4-10)$ $=+2.24 \times 10^{-16} {\text{J}}$