Electrostatics

The forces acting on the oil drop are its weight, buoyant force, and electrostatic force. The buoyant force on the oil drop is very small as compared to other two forces. Thus, the weight of the spherical oil drop is balanced by the electrostatic force

$qE = mg$

$qE = {4 \over 3}\pi {r^3}\rho g$ ..... (i)

where, r = radius of oil drop, $\rho$ = density of oil, q = charge on the oil drop

IN the absence of electric field,

IN equilibrium,

Viscous force on the drop = Weight of the drop

$6\pi \eta r{v_T} = mg$

$6\pi \eta r{v_T} = {4 \over 3}\pi {r^3}\rho g$ ...... (ii)

where, v_T = terminal velocity, $\eta $ = coefficient of viscosity of air

or ${r^2} = {{18 \times \eta \times {v_T}} \over {4 \times \rho \times g}}$

Substituting the given values, we get

${r^2} = {{18 \times 1.8 \times {{10}^{ - 5}} \times 2 \times {{10}^{ - 3}}} \over {4 \times 900 \times 9.8}}$ or $r = {3 \over 7} \times {10^{ - 5}}$ m

From equation (i), we get $q = {{4\pi {r^3}\rho g} \over {3E}}$

Substituting the given values, we get

$q = {{4 \times \pi \times {{\left( {{3 \over 7} \times {{10}^{ - 5}}} \right)}^3} \times 900 \times 9.8} \over {3 \times \left( {{{81\pi } \over 7} \times {{10}^5}} \right)}}$

$ = {{4 \times \pi \times 7 \times 3 \times 3 \times 3 \times {{10}^{ - 15}} \times 900 \times 9.8} \over {3 \times 81\pi \times {{10}^5} \times 7 \times 7 \times 7}} = 8 \times {10^{ - 19}}C$

Electrostatic pressure is

$\left( {{{{\sigma ^2}} \over {2{\varepsilon _0}}}} \right)$ N/m²

The force is

$F = {{{\sigma ^2}} \over {2{\varepsilon _0}}} \times \pi {R^2}$

Therefore, $F \propto {{{\sigma ^2}{R^2}} \over {{\varepsilon _0}}}$

Number of electric field lines originating from Q₁ is more than terminating at Q₂.

$\therefore$ $|{Q_1}| > |{Q_2}|$

Here, Q₁ is positive while Q₂ is negative.

The electric field at a distance x towards the right of Q₂ is given by

$|\overrightarrow E | = {1 \over {4\pi { \in _0}}}{{{Q_1}} \over {{{(d + x)}^2}}} - {1 \over {4\pi { \in _0}}}{{{Q_2}} \over {{x^2}}}$,

where d is the separation between Q₁ and Q₂. Since ${Q_1} > {Q_2}$, $|\overrightarrow E |$ becomes zero for some finite x.

According to Gauss's law, we have

$\mathrm{Flux=\frac{Charge~enclosed}{\varepsilon_0}}$

The enclosed charges are half of the charges in the disc, point charge at $(a/4,-a/4,0)$, and charge in the rod from point $a/4$ to $a/2$, that is, 8 C/4. Therefore,

Flux $ = {{3C - 7C + (8C/4)} \over {{\varepsilon _0}}} = - {{2C} \over {{\varepsilon _0}}}$

Surface charge density is

$\rho = {Q \over {4\pi {R^2}}}$

where R is the radius of the sphere. For cancelling the field inside the cavity due to the shell of radius R, the shell with radius 2R induces a charge $-$Q in its inner surface and hence the total charge on the outer surface is $Q_1+Q_2$. Similarly, the charge on the outermost shell is

${Q_1} + {Q_2} + {Q_3}$

It is given that the surface charge densities $\rho_1,\rho_2$ and $\rho_3$ are equal, that is,

${{{Q_1}} \over {4\pi {R^2}}} = {{{Q_1} + {Q_2}} \over {4\pi {{(2R)}^2}}} = {{{Q_1} + {Q_2} + {Q_3}} \over {4\pi {{(3R)}^2}}}$

$ \Rightarrow {Q_1} = {{{Q_1} + {Q_2}} \over 4} = {{{Q_1} + {Q_2} + {Q_3}} \over 9}$

That is,

${Q_1} = {{{Q_1}} \over 4} + {{{Q_2}} \over 4}$

$ \Rightarrow {Q_2} = 3{Q_1}$

Therefore,

${Q_3} = 5{Q_1}$

Hence, the ratio of the charges given to the shells ${Q_1}:{Q_2}:{Q_3}$ is 1 : 3 : 5.

Case (P) : $E=0,V=0, B=0,\mu=0$.

Case (Q) : $E\ne0,V=0, B=0,\mu=0$.

Case (R) : $E=0,V\ne0, B\ne0,\mu\ne0$.

Case (S) : $E=0,V\ne0, B\ne0,\mu\ne0$.

Case (T) : $E\ne0,V=0, B=0,\mu=0$.

Electric field at point O is vector sum of electric field due to all three point charges individually.

$\overrightarrow E = \overrightarrow {{E_A}} + \overrightarrow {{E_B}} + \overrightarrow {{E_C}} $ and $\overrightarrow {{E_A}} = - \overrightarrow {{E_B}} $

So, $\overrightarrow E = \overrightarrow {{E_C}} = {{2q/3} \over {4\pi {\varepsilon _0}.{R^2}}}$ along $\overrightarrow {OC} $

$ = {q \over {6\pi \varepsilon _0^2.{R^2}}}$ along $\overrightarrow {OC} $

$ = {{ + q'} \over {6\pi \varepsilon _0^2.{R^2}}}\widehat i$

$ = {{ - q'} \over {6\pi \varepsilon _0.{R^2}}}\widehat j$

From geometry, we can find that $\angle$ABC = 30$^\circ$ and $\angle$ACB = 90$^\circ$

So, AB = 2R, AC = R, BC = $\sqrt3$R

Total potential energy of the system is,

$U = {1 \over {4\pi {\varepsilon _0}}}\left[ {{{q/3 \times q/3} \over {2R}} - {{q/3 \times 2q/3} \over R} - {{q/3 \times 2q/3} \over {\sqrt 3 R}}} \right] \ne 0$

$F = {{q/3 \times q/3} \over {4\pi {\varepsilon _0} \times {{(\sqrt 3 R)}^2}}} = {{{q^2}} \over {54\pi {\varepsilon _0}{R^2}}}$

Potential at point O is,

$V = {{q/3 \times q/3 - 2q/3} \over {4\pi {\varepsilon _0}R}} = 0$

Time constant ($\tau$) = RC$_{eq}$

Where R = Resistance

C$_{eq}$ = equivalent capacitance

And, $v = {{ - dx} \over {dt}} \Rightarrow dx = - vdt$

$ \Rightarrow \int_{d/3}^x {dx = - v\int_0^t {dt} } $

$ \Rightarrow [x]_{d/3}^x = - v[t]_0^t$

$x - {d \over 3} = - vt$

$x = {d \over 3} - vt$ ..... (i)

And also, ${1 \over {{C_{eq}}}} = {1 \over {{C_1}}} + {1 \over {{C_2}}} = {1 \over {{{{\varepsilon _0}A} \over {(d - x)}}}} + {1 \over {{{{\varepsilon _0}Ak} \over x}}}$

${1 \over {{C_{eq}}}} = {{d - x} \over {{\varepsilon _0}A}} + {x \over {{\varepsilon _0}Ak}}$

$ = {1 \over {{\varepsilon _0}A}}\left[ {d - x + {x \over k}} \right]$

$ = {1 \over {{\varepsilon _0}A}}\left[ {{{(d - x)k + x} \over k}} \right]$

${1 \over {{C_{eq}}}} = {1 \over {{\varepsilon _0}Ak}}[dk - xk + x]$

$\therefore$ ${C_{eq}} = {{{\varepsilon _0}A} \over {kd + x(1 - k)}}$ .... (ii)

Given, $A = 1,k = 2,x = {d \over 3} - vt$

Hence, ${C_{eq}} = {{{\varepsilon _0} \times 1} \over {2d + \left[ {{d \over 3} - vt} \right](1 - 2)}}$

$ = {{6{\varepsilon _0}} \over {5d + 3vt}}$

Therefore, $\tau = R{C_{eq}} = {{6R{\varepsilon _0}} \over {5d + 3vt}}$

Statement 1 is True.

The earth is used as a reference with zero potential. The reasons are :

(i) Due to large size of the earth, its potential does not change even if a small amount of charge is given to the earth or taken away from it. Also, all conductors on the earth which are not given any external charge, are very nearly at the potential V_e. The choice of reference is quite arbitrary and is largely governed by convenience.

(ii) The earth is a good conductor. It is easily accessible for electrical circuits geographically distributed over it and hence it is used as a common reference with zero potentil.

Statement 2 is true.

V = ${Q \over {4\pi {\varepsilon _0}R}}$

Therefore, statement 2 is correct explanation for statement 1.

At $r = R$,

By applying Gauss law,

$\phi = {{{q_{enclosed}}} \over {{\varepsilon _0}}}$

$ \Rightarrow \oint {E.ds = {{{q_{enclosed}}} \over {{\varepsilon _0}}}} $

$ \Rightarrow E\oint {ds = {{{q_{enclosed}}} \over {{\varepsilon _0}}}} $

$ \Rightarrow E.4\pi {R^2} = {{Ze} \over {{\varepsilon _0}}}$

$ \Rightarrow E = {1 \over {4\pi {\varepsilon _0}}}{{Ze} \over {{R^2}}}$

i.e., $E\,\mu {1 \over {{R^2}}}$

Here, electric field is independent of 'a'.

For a = 0, the graph is as shown, the equation for the graph line is

The charge in the dotted element shown in figure is

$dq = \sigma \times 4\pi {r^2}dr$

$\therefore$ $dq = \left( {d - {d \over R}r} \right)4\pi {r^2}dr$

$ \Rightarrow Ze = \int_0^R {4\pi d{r^2}dr - \int_0^R {{{4\pi d} \over R}{r^3}dr} } $

$Ze = 4\pi d{{{R^3}} \over 3} - {{4\pi d} \over R}{{{R^4}} \over 4}$

$\therefore$ ${{Ze} \over {4\pi d{R^3}}} = {1 \over 3} - {1 \over 4} = {1 \over {12}}$

$\therefore$ $d = {{3Ze} \over {\pi {R^3}}}$

The electric field inside a uniformly charged sphere is proportional to r. The volume charge density inside a uniformly charged sphere is constant and hence a = R.

When cavity is not present:

Electric field inside the solid sphere

$ \overrightarrow{\mathrm{E}_{1}}=\frac{\rho}{3 \varepsilon_{0}} \overrightarrow{r_{1}} $ ..... (i)

Solid sphere after making cavity inside solid sphere,

$ \overrightarrow{\mathrm{E}_{2}}=\frac{\rho}{3 \varepsilon_{0}} \overrightarrow{r_{2}} $ ...... (ii)

${\overrightarrow E _{net}} = \overrightarrow {{E_1}} - \overrightarrow {{E_2}} = {\rho \over {3{\varepsilon _0}}}\overrightarrow {{r_1}} - {\rho \over {3{\varepsilon _0}}}\overrightarrow {{r_2}} $

$=\frac{\rho}{3 \varepsilon_{0}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right) $ ...... (iii)

From eq (iii), we get

$\therefore \quad \mathrm{E}_{\mathrm{net}}=\frac{\rho}{3 \varepsilon_{0}}(\vec{r})$

Hence, electric field inside the emptied space (cavity) is non-zero and uniform.

Consider that,

$\mathrm{P}=(-a, 0,0)$ and $\mathrm{Q}=(0, a, 0)$ points lie on the equatorial plane of dipole.

The charges makes an electric dipole.

Therefore, potential at $\mathrm{P}=$ Potential at $\mathrm{Q}=0$

Work done $(\mathrm{W})=q\left(\mathrm{~V}_{\mathrm{P}}-\mathrm{V}_{\mathrm{Q}}\right)=0$

Let the radii of the inner and outer cylinders be $r_1$ and $r_2$, respectively. The problem has cylindrical symmetry and Gauss's law can be applied to find the electric field $E$.

In the case (A), surface charge density $\sigma$ is given to the inner cylinder. Take a cylinder of radius $r\left(r_1<\right. r

$ E(2 \pi r l)=\frac{q_{\mathrm{enc}}}{\epsilon_0}=\frac{\sigma\left(2 \pi r_1 l\right)}{\epsilon_0}, \quad \text { i.e., } \quad E=\frac{\sigma r_1}{\epsilon_0 r} . $

Potential difference between the two cylinders is given by

$ \Delta V=-\int_{r_1}^{r_2} E \mathrm{~d} r=-\frac{\sigma r_1}{\epsilon_0} \ln \left(\frac{r_2}{r_1}\right) $

In the case (B), charge density $\sigma$ is given to the outer cylinder. Apply Gauss's law to get $E=0$ (because $q_{\mathrm{enc}}=0$ ) and hence $\Delta V=0$.

In the case (C), uniform linear charge density $\lambda$ is kept along the common axis of cylinders. Apply Gauss's law as in case (A) to get the electric field

$ E(2 \pi r l)=\lambda l / \epsilon_0, \quad \text { i.e., } \quad E=\lambda /\left(2 \pi \epsilon_0 r\right) . $

Potential difference between the two cylinders is

$ \Delta V=-\int_{r_1}^{r_2} E \mathrm{~d} r=-\frac{\lambda}{2 \pi \epsilon_0} \ln \left(\frac{r_2}{r_1}\right) $

In the case (D) charge density $\sigma$ is given to both the cylinders. Using the argument similar to the case (A), we get, $\Delta V=-\frac{\sigma r_1}{\epsilon_0} \ln \left(\frac{r_2}{r_1}\right)$.

Net charge on sphere = $-20+20$ = 0

When we brought +q charge near neutral conducting sphere induction no produced.

So, net charge on sphere should be zero.

When +q is placed option (d).

Outside the conducting sphere, rearrange of charge taken place on sphere.

(A) Since, potential is same, so electrostatic energy stored is zero.

(B)

$ \text { (C) There is no charge in air, } \therefore r=\mathrm{R}_0 $

(D) From $r=\mathrm{R}_0$ to $\infty$.

We can assume point charge at centre equivalent to $q$.

$ \text { So, } \overrightarrow{\mathrm{E}} \propto \frac{1}{r^2} \text { so, it is discontinuous } $

A liquid-bubble contains air inside and liquid in the outer surface. When the bubble collapses into a droplet, the volume of liquid remains same in this transformation.

Let thickness of liquid in surface = $t$

$\therefore$ Volume of liquid in bubble $=4\pi a^2t$

Volume of the droplet $=\frac{4}{3}\pi R^3$

$\therefore~\frac{4}{3}\pi R^3=4\pi a^2t$

$R=(3a^2t)^{1/3}$ ..... (i)

Let the potential of liquid-bubble = V

$\therefore \mathrm{V}=\frac{+q}{4 \pi \varepsilon_{0}}$ where $q$ denotes charge on the bubble.

$\therefore$ Potential of droplet $ = {2 \over {4\pi {\varepsilon _0}R}}$

Potential of droplet $ = \left( {{{4\pi {\varepsilon _0}aV} \over {4\pi {\varepsilon _0}}}} \right)\left[ {{1 \over {{{(3{a^2}t)}^{1/3}}}}} \right]$

Potential of droplet $ = V\left[ {{a \over {{{(3{a^2}t)}^{1/3}}}}} \right]$

Potential of droplet $ = V{\left[ {{{{a^3}} \over {3{a^2}t}}} \right]^{1/3}}$

Potential of droplet $ = V{\left[ {{a \over {3t}}} \right]^{1/3}}$