Electrostatics Practice Questions

$\tan 37°=\frac{3}{4}={\frac{{kP}_{2}}{{r}^{3}}}/{\frac{2 {kP}_{1}}{{r}^{3}}}=\frac{{P}_{2}}{2 {P}_{1}}=\frac{3}{4}$ $\frac{{P}_{2}}{{P}_{1}}=\frac{3}{2}$ $\frac{{P}_{1}}{{P}_{2}}=\frac{2}{3}$

${E}_{1}=\frac{{kq}}{l^{2}}={E}_{2}$ ${E}_{3}=\frac{{kq}}{(\\sqrt{2} l)^{2}}=\frac{{kq}}{2 l^{2}}$ ${E}=\frac{\\sqrt{2} {kq}}{l^{2}}-\frac{{kq}}{2 l^{2}}=\frac{{kq}}{2 l^{2}}(2 \\sqrt{2}-1)$

In the given statements (a), (b) and (c) are correct. Only (d) is wrong. Since, if electric field lines of force (or electric field) is parallel to Gaussian surface, then angle between normal to surface and electric field is θ = 90°. ∴ By equation of electric flux ϕ = EA cos θ = EA cos 90° = 0 Hence, electric flux will be zero in this situation. Hence, option (c) is the correct.

Given, the radius of oil drop, $r=2 {mm}=0.002 {m}$ The density of oil drop, $\rho =3 {g} / {cm}^{3}=3 \times 10^{3} {kg} / {m}^{3}$ The constant electric field, $E=3.55 \times 10^{5} {Vm}^{-1}$ The constant electric field, $E=3.55 \times 10^{5} {Vm}^{-1}$ Under stationary condition of oil drop, $m g=q E$ $\Rightarrow \;\; \rho V g=q E$ $[ \because m=\rho V]$ $\Rightarrow \;\; \rho (\;\frac{4}{3} \pi r^{3}) g=n e E$ $[ \because V=\;\frac{4}{3} \pi r^{3}$ and $q=n e]$ $n=\;\frac{\rho (\frac{4}{3} \pi r^{3}) g}{e E}$ $\Rightarrow \;\; n=\;\frac{3 \times 10^{3} \times (\frac{4}{3} \pi (0.002)^{3}) 9.81}{1.6 \times 10^{-19} \times (3.55 \times 10^{5})}$ $\Rightarrow \;\; n=1.73 \times 10^{10}$

Given, mass of block, $m=1 {kg}$ Acceleration due to gravity, $g=9.8 {ms}^{-2}$ Inclination, $\theta =30^{\circ}$ Electric field, $E=200 {N} / {C}$ Coefficient of friction, $\mu =0.2$ Charge, $q=5 {mC}=5 \times 10^{-3} {C}$ Let friction force, $f=\mu N$ where, $N$ be the normal reaction. Since, net force is zero along the perpendicular direction of incline. Therefore, force along $Y$-axis will be zero. $\Rightarrow \;\; N \;\; =m g \cos 30^{\circ}+q E \sin 30^{\circ}$ $\Rightarrow \;\; N \;\; =1 \times 9.8 \times \sqrt{3} / 2+5 \times 10^{-3} \times 200 \times 1 / 2$ $\;\; =8.49+0.5$ $\;\; =8.99 {N} ∼eq 9 {N}$ $\therefore \;\; f=\mu N=\;\frac{2}{10} \times 9=\;\frac{18}{10}=1.8 {N}$ Now, total force along the plane of incline, $\;m g \sin 30^{\circ}-f-q E \cos 30^{\circ}=m a\;$ $\Rightarrow \;1 \times 9.8 \times 1 / 2-1.8-5 \times 10^{-3} \times 200 \sqrt{3} / 2\;=a$ $\Rightarrow \;5-1.8-1.732 / 2\;=a$ $\Rightarrow \;a=2.34 {ms}^{-2}$ Since, initial velocity of body, $u=0 {ms}^{-1}$ and distance along incline, $s=h / \sin 30^{\circ}=1 / \sin 30^{\circ}=2$ By using second equation of motion, $\;s \;\; =u t+1 / 2 {at}^{2}$ $\Rightarrow \;2 \;\; =0+1 / 2 \times 2.34 \times t^{2}$ $\Rightarrow \;t^{2} \;\; =\;\frac{4}{2.34}$ $\Rightarrow \;t=\;\frac{2}{\sqrt{2.34}}$

The $1 {C}$ charge present in the origin and other $1 \mu {C}$ charge are placed at $1 {m}, 2 {m}, 4 {m}, 8 {m} . . .$Using the Coulomb's law, $F=\;\frac{k q_{1} q_{2}}{r^{2}}$ $F=9 \times 10^{9} \times (1) \times 10^{-6} \;\; [1+\;\frac{1}{2^{2}}+\;\frac{1}{4^{2}}+\;\frac{1}{8^{2}}+⋯]$ $=9 \times 10^{3} \times [\;\frac{1}{1-\;\frac{1}{4}}]=12 \times 10^{3} {N}$ The value of $x$ to the nearest integer is 12 .

Electric field due to each sheet is uniform and equal to $E=\sigma /{2\epsilon _0}$Now net electric field between plates$\vec{E}_{\text{net}}=E\cos60°(-\hat{x})$ $+(E-E\sin60°)(\hat{y})$ $=\sigma /{2\epsilon _0}[-\hat{x}/2+(1-\sqrt3/2)\hat{y}]$

We solve this using the principle of superposition. The electric field of the sphere with a cavity is the field of the complete sphere (charge density $\rho$) plus the field of a sphere at the cavity's location with opposite charge density ($-\rho$). Let the center of the large sphere be the origin.

At point A (center of the cavity, at $r=R/2$), the field from the large sphere is $E_{large} = \frac{\rho r}{3\epsilon_0} = \frac{\rho (R/2)}{3\epsilon_0}$. The field from the superimposed negative sphere is zero at its own center. Therefore, $|E_A| = \frac{\rho R}{6\epsilon_0}$.

At point B (on the surface, at $r=-R$), the field from the large sphere is $|E_{large}| = \frac{\rho R}{3\epsilon_0}$ (downwards). The field from the $-\rho$ sphere (center at $r=R/2$, radius $R/2$) at point B is at a distance of $d=3R/2$. Its field is $|E_{small}| = \frac{k|Q_{small}|}{d^2} = \frac{1}{4\pi\epsilon_0}\frac{\rho \frac{4}{3}\pi(R/2)^3}{(3R/2)^2} = \frac{\rho R}{54\epsilon_0}$ (upwards).

The net field at B is the vector sum, so we subtract the magnitudes: $|E_B| = |E_{large}| - |E_{small}| = \frac{\rho R}{3\epsilon_0} - \frac{\rho R}{54\epsilon_0} = \frac{17\rho R}{54\epsilon_0}$.

The ratio of the magnitudes is $\frac{|E_A|}{|E_B|} = \frac{\rho R / (6\epsilon_0)}{17\rho R / (54\epsilon_0)} = \frac{1}{6} \times \frac{54}{17} = \frac{9}{17} = \frac{18}{34}$.

$v^2 = u^2 + 2as$ $v^2=0+2({qE}/m)x$ $v^2={2qE}/mx$

The flux passes through ABCD (x - y) plane is zero, because electric field parallel to surface. Flux of the electric field through surface BCGF (y - z) At BCGF (electric field) $\Rightarrow E ^{\to }= 12\hat{i}-(y^{2+}1)\hat{j}$(x = 3m)Flux $ϕ_11$ = 12 × 4 = 48 $Nm^2$/CSo $ϕ_1-ϕ_11$ = 0 - 48 = - 48 $Nm^2$/C∴ Correct answer -48

$E_x={K(4q)}/R^2\cos30°$ $+{K(2q)}/R^2\cos30°$ $+{K(2q)}/R^2\cos30°$

${\text{E}}_{2}=$ electric field due to ${\text{Q}}_{2}$ $={{\text{kQ}}_{2}}/{{x}_{2}^{2}}$ ${\text{E}}_{1}={{\text{kQ}}_{1}}/{{x}_{1}^{2}}$ From diagram, $\tan \theta ={{\text{E}}_{2}}{{\text{E}}_{1}}=\frac{{x}_{1}}{{x}_{2}}$ ${{\text{kQ}}_{2}}/{{x}_{2}^{2} \times {{\text{kQ}}_{1}}/{{x}_{1}^{2}}}=\frac{{x}_{1}}{{x}_{2}}$ ${{\text{Q}}_{2} {x}_{1}^{2}}/{{\text{Q}}_{1} {x}_{2}^{2}}=\frac{{x}_{1}}{{x}_{2}}$ $\frac{\text{Q}_{1}}{\text{Q}_{2}}=\frac{x_{1}}{x_{2}}$

Inside the shell, $E=0$ hence ${F}=0$ Outside the shell, ${\text{E}}=\frac{1}{4 \pi \epsilon _{0}} {{\text{Q}}}/{{r}^{2}}$ hence ${\text{F}}=\frac{1}{4 \pi \epsilon _{0}} {{\text{Qq}}}/{{r}^{2}}$ for ${r} > {\text{R}}$

Potential of centre $= {V}=$ $∑(\frac{{kq}}{ {R}} )$ ${V}_{ {C}}=\frac{{K}(\sum {q})}{\ {R}}$ ${V}_{{C}}=\frac{{K}(0)}{ {R}}=0$ Electric field at centre ${\vec{E}}_{ {B}}=\sum {\vec{E}}$ Let $\mathrm{E}$ be electric field produced by each charge at the centre, then resultant electric field will be${E}_{ {C}}=0,$ since equal electric field vectors are acting at equal angle so their resultant is equal to zero.

${E}=\frac{{KQ}_{1}}{{r}^{2}}$ $\Delta{V}=\int _{ {R}} ^{4 {R}} {E} {dr}= \frac{3 \ {KQ}_{1}}{4 \ {R}}$

Since $\vec{r}$ and $\vec{p}$ are perpendicular to each other therefore point lies on the equitorial plane. Therefore electric field at the point will be antiparallel to the dipole moment.i.e. $\vec{E}||-\vec{p}$ $\vec{E}||(\hat{i}+3\hat{j}-2\hat{k})$

since initial velocity is zero and acceleration of particle will be constant, so particle willtravel on a straight line path.

${E}= {E}_{0} (1- {ax}^{2} )$ ${W}=∫{qE} {dx}$ $= {qE}_{0} ∫_ {0} ^{ {x}_{0}} (1- {ax}^{2} ) {dx}$ $= {qE}_{0}[{x}_{0}-\frac{{ax}_{0}^{3}}{3} ]$ For $\Delta{KE}=0, \;\;{W}=0$ Hence ${x}_{0}=\sqrt{\frac{3}{\ {a}}}$