Electrostatics

Place a point charge $q$ at the centre $P$ of a cylindrical Gaussian surface. Because of symmetry, the electric flux across each circular end cap is the same; call this $\phi_f$. Let $\phi$ be the flux through the curved side. Applying Gauss’s law to the closed cylinder, the total flux is

$ \phi_{\text{total}} \;=\; 2\phi_f \;+\; \phi \;=\; \frac{q}{\varepsilon_0}. $

The solid angle subtended at point $P$ by the flat surface (spherical cap of half-angle $\theta$ ) is $\Omega_f=2 \pi(1-\cos \theta$ ). The total solid angle is $\Omega_t=4 \pi$. Thus, the flux through one flat surface is

$ \phi_f=\phi_t\left(\frac{\Omega_f}{\Omega_t}\right)=\frac{\phi_t(1-\cos \theta)}{2} $

Solve the above equations to get

$ \phi=\frac{q \cos \theta}{\epsilon_0}=\frac{q \cos 30^{\circ}}{\epsilon_0}=\frac{\sqrt{3} q}{2 \epsilon_0} $

The flux through the curved surface for $\theta=60^{\circ}$ is

$ \phi^{\prime}=\frac{q \cos 60^{\circ}}{\epsilon_0}=\frac{q}{2 \epsilon_0}=\frac{\phi}{\sqrt{3}} $

due to wire

$\begin{aligned} & \mathrm{dV}=-\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dx}} \\ & \int_\limits{\mathrm{v}_{\mathrm{P}}}^{\mathrm{v}_{\mathrm{R}}} \mathrm{dV}=-\int_\limits{0.5}^2 \frac{2 \mathrm{k} \lambda}{\mathrm{x}} \mathrm{dx} \\ & \mathrm{v}_{\mathrm{R}}-\mathrm{v}_{\mathrm{P}}=-2 \mathrm{k} \lambda \ln \frac{2}{0.5} \\ & =-2 \times 9 \times 10^9 \times 3 \times 10^{-9} \times 2 \times 0.7=-126 \mathrm{~V} \end{aligned}$

due to sphere

$\begin{aligned} & v_R-v_P= \frac{k Q}{2}-\frac{k Q}{1}=-\frac{k Q}{2}=\frac{-9 \times 10^9 \times 10 \times 10^{-9}}{2} \\ &=-45 \mathrm{~V} \\ & v_R-v_P=-126-45=-171 \mathrm{~V} \\ & v_P-v_R= 171 \mathrm{~V} \end{aligned}$

Sol. From Gauss law,

$ \phi_{\text {hemisphere }}+\phi_{\text {Cone }}=\frac{\mathrm{q}}{\varepsilon_0} $ ............(1)

Total flux produced from $\mathrm{q}$ in $\alpha$ angle

$ \phi=\frac{\mathrm{q}}{2 \varepsilon_0}[1-\cos \alpha] $

For hemisphere, $\alpha=\frac{\pi}{2}$

$ \phi_{\text {hemisphere }}=\frac{\mathrm{q}}{2 \varepsilon_0} $

From equation (1)

$ \begin{aligned} & =\frac{\mathrm{q}}{2 \varepsilon_0}+\phi_{\text {cone }}=\frac{\mathrm{q}}{\varepsilon_0} \\\\ & \phi_{\text {cone }}=\frac{\mathrm{q}}{2 \varepsilon_0} \\\\ & \frac{4 \mathrm{q}}{6 \varepsilon_0}=\frac{\mathrm{q}}{2 \varepsilon_0} \\\\ & \mathrm{n}=3 \end{aligned} $

V at B is zero if

${{kQ} \over {(2 + R + x)}} = {{{{kQ} \over {\sqrt 3 }}} \over {x + R}}$ ($\because$ $k = {1 \over {4\pi {\varepsilon _0}}}$)

$\sqrt 3 (x + R) = 2 + R + x$

$(\sqrt 3 - 1)x + (\sqrt 3 - 1)R = 2$ .....(i)

V at A is zero if

${{kQ} \over {2 - x'}} = {{{{kQ} \over {\sqrt 3 }}} \over {x'}}$

$\sqrt 3 x' = 2 - x'$

$x' = {2 \over {\sqrt 3 + 1}}$

$x' + x = R$

${2 \over {\sqrt 3 + 1}} + x = R$

$2 + (\sqrt 3 + 1)x = (\sqrt 3 + 1)R$

$x = {{(\sqrt 3 + 1)R - 2} \over {\sqrt 3 + 1}}$

$(\sqrt 3 + 1)R - (\sqrt 3 + 1)x = 2$ .....(ii)

Using Eqs. (i) and (ii), we get

$R = \sqrt 3 m = 1.73m$

V at B is zero if

${{kQ} \over {(2 + R + x)}} = {{{{kQ} \over {\sqrt 3 }}} \over {x + R}}$ ($\because$ $k = {1 \over {4\pi {\varepsilon _0}}}$)

$\sqrt 3 (x + R) = 2 + R + x$

$(\sqrt 3 - 1)x + (\sqrt 3 - 1)R = 2$ .....(i)

V at A is zero if

${{kQ} \over {2 - x'}} = {{{{kQ} \over {\sqrt 3 }}} \over {x'}}$

$\sqrt 3 x' = 2 - x'$

$x' = {2 \over {\sqrt 3 + 1}}$

$x' + x = R$

${2 \over {\sqrt 3 + 1}} + x = R$

$2 + (\sqrt 3 + 1)x = (\sqrt 3 + 1)R$

$x = {{(\sqrt 3 + 1)R - 2} \over {\sqrt 3 + 1}}$

$(\sqrt 3 + 1)R - (\sqrt 3 + 1)x = 2$ .....(ii)

Using Eqs. (i) and (ii), we get

$R = \sqrt 3 m = 1.73m$

$x = {{(\sqrt 3 + 1)\sqrt 3 - 2} \over {\sqrt 3 + 1}} = {{\sqrt 3 + 1} \over {\sqrt 3 + 1}} = 1$ m

Hence, the centre of circle is having x-coordinate

= b = 2 + x = 3.00 m.

U_i = 0

${U_f} = {{kqp} \over {{{\left( {2l\sin {\alpha \over 2}} \right)}^2}}}$ + mgh .... (i)

Now, from $\Delta$OAB,

$\alpha$ + 90 $-$ $\theta$ + 90 $-$ $\theta$ = 180

$\Rightarrow$ $\alpha$ = 2$\theta$

From $\Delta$ABC, h = 2l sin$\left( {{\alpha \over 2}} \right)$ sin$\theta$

h = 2l sin$\left( {{\alpha \over 2}} \right)$ sin$\left( {{\alpha \over 2}} \right)$ $\Rightarrow$ h = 2l sin²$\left( {{\alpha \over 2}} \right)$

Now, charge is in equilibrium at point B.

So, using sine rule,

${{mg} \over {\sin \left( {90 + {\alpha \over 2}} \right)}} = {{qE} \over {\sin 1(80 - 2\theta )}}$

$ \Rightarrow {{mg} \over {\cos {\alpha \over 2}}} = {{qE} \over {\sin 2\theta }}$

$ \Rightarrow {{mg} \over {\cos {\alpha \over 2}}} = {{qE} \over {\sin \alpha }} \Rightarrow {{mg} \over {\cos {\alpha \over 2}}} = {{qE} \over {2\sin {\alpha \over 2}\cos {\alpha \over 2}}}$

$ \Rightarrow qE = mg2\sin \left( {{\alpha \over 2}} \right)$

$ \Rightarrow {{q2kp} \over {{{\left( {2l\sin {\alpha \over 2}} \right)}^3}}} = mg2\sin \left( {{\alpha \over 2}} \right)$

$ \Rightarrow {{kpq} \over {{{\left( {2l\sin {\alpha \over 2}} \right)}^2}}} = mg\sin \left( {{\alpha \over 2}} \right) \times \left( {2l\sin {\alpha \over 2}} \right)$

$ \Rightarrow {{kpq} \over {{{\left( {2l\sin {\alpha \over 2}} \right)}^2}}} = mgh$

Substituting this in Eq. (i), we get

${U_f} = mgh + {{kpq} \over {{{\left( {2l\sin {\alpha \over 2}} \right)}^2}}} \Rightarrow {U_f} = 2mgh$

W = $\Delta$U = 2mgh = Nmgh

$ \therefore $ N = 2

$qE = Mg $

$\Rightarrow q = {{mg} \over E} = \left( {{{mg} \over {V/d}}} \right)$

$ = {{900 \times {4 \over 5}\pi (8 \times {{10}^{ - 7}}) \times 10} \over {{{200} \over {0.01}}}}$

$N = {q \over e} = {{900 \times 4\pi \times {8^3} \times {{10}^{ - 21}} \times 10 \times 0.01} \over {200 \times 1.6 \times {{10}^{ - 19}}}}$

N = 6

To determine the disc’s total charge, integrate the radial surface–charge distribution across its entire area:

$ \begin{aligned} q_{\text{enc}} &= \int_{0}^{a} \sigma(r)\, 2\pi r\,dr = 2\pi \sigma_{0}\int_{0}^{a}\!\left(r - \frac{r^{2}}{a}\right)dr \\[4pt] &= 2\pi \sigma_{0}\!\left[\frac{r^{2}}{2} - \frac{r^{3}}{3a}\right]_{0}^{a} = \frac{\pi \sigma_{0}a^{2}}{3}. \end{aligned} $

The electric flux through the spherical surface that encloses the entire charged disc is

$ \phi_0=\frac{q_{\mathrm{enc} 0}}{\epsilon_0}=\frac{\pi \sigma_0 a^2}{3 \epsilon_0} . $

Next, we find the flux through the smaller spherical surface of radius $a / 4$, which encloses a portion of the disc. The charge on this portion of the disc is

$ q_{\mathrm{enc}}=2 \pi \sigma_0 \int\limits_0^{a / 4}\left(r-\frac{r^2}{a}\right) \mathrm{d} r=\frac{5}{96} \pi \sigma_0 a^2 $

The flux through this spherical surface is

$ \phi=\frac{q_{\mathrm{enc}}}{\epsilon_0}=\frac{5 \pi \sigma_0 a^2}{96 \epsilon_0} $

The ratio of the two fluxes is

$ \frac{\phi_0}{\phi}=\frac{32}{5}=6.4 $

Electrostatic force = Spring force

At l, $Fe = {F_{sp}}kl = {{2\alpha pq} \over {{l^3}}}$ (Here, $\alpha = {1 \over {4\pi {\varepsilon _0}}}$)



Now, the mass m is displaced by $\Delta $l = x from the mean position.

${F_{net}} = {F_{sp}} - Fe = k(l + x) - {{q(2\alpha p)} \over {{{(l + x)}^3}}}$

$ = k(x + 1) - {{q(2\alpha p)} \over {{l^3}{{(l + x/l)}^3}}}$

$ = kx + kl - q\left( {{{2\alpha p} \over {{l^3}}}} \right)\left( {1 - {{3x} \over l}} \right)$

$ = kx + kl - q\left( {{{2\alpha p} \over {{l^3}}}} \right) + {{2\alpha p} \over {{l^3}}}.{{3x} \over l}$

Substituting ${{2\alpha pq} \over {{l^3}}} = kl$, we get

${F_{net}} = kx + kl\left( {{{3x} \over l}} \right) = 4kx$

This is restoring in nature.

Hence, ${k_{eq}} = 4k$

or $T = 2\pi \sqrt {{m \over {4k}}} = \pi \sqrt {{m \over k}} $

$ \therefore $ $f = {1 \over \pi }\sqrt {{k \over m}} $

So, $\delta = \pi = 3.14$

Given, mass of particle = 10^$-$3 kg, charge on particle = 1.0 C, electric field $\overrightarrow E (t) = {E_0}\sin \omega t\,\widehat i$, E₀ = 1.0 N C^$-$1, $\omega$ = 10³ rad s^$-$1

Force on particle is given by

$ \Rightarrow \overrightarrow F = q{E_0}\sin \omega t\,\widehat i = 1.0 \times 1.0 \times \sin ({10^3}t)\widehat i$

$ \Rightarrow \overrightarrow F = \sin ({10^3}t)\widehat i$

We know that $\overrightarrow F = m\overrightarrow a \Rightarrow \overrightarrow a = {{\overrightarrow F } \over m}$

$ \Rightarrow a = {{\sin ({{10}^3}t)} \over {{{10}^{ - 3}}}} \Rightarrow a = {10^3}\sin ({10^3}t)$

We know $a = {{dv} \over {dt}}$

$ \Rightarrow {{dv} \over {dt}} = {10^3}\sin ({10^3}t)$

$ \Rightarrow dv = {10^3}\sin ({10^3}t)dt$

Now, integrating it, we get

$\int\limits_0^v {dv = \int\limits_0^t {{{10}^3}\sin ({{10}^3}t)dt} } $

$ \Rightarrow \left. v \right|_0^v = {10^3}\left( {{{ - \left. {\cos ({{10}^3}t)} \right|_0^t} \over {{{10}^3}}}} \right) \Rightarrow \left. v \right|_0^v = \left. { - \cos ({{10}^3}t)} \right|_0^t$

$ \Rightarrow v = ( - \cos {10^3}t + \cos 0)$

$ \Rightarrow v = (1 - \cos ({10^3}t))$

We know that cos$\theta$ can take values between $-$1 and 1. Therefore, maximum speed attained by the particle when cos$\theta$ = $-$1. Thus,

${v_{\max }} = 1 - ( - 1) = 2$ m s^$-$1

ANBP is cross-section of a cylinder of length L. The line charge passes through the centre O and perpendicular to paper.

$AM = {a \over 2}$, $MO = {{\sqrt 3 a} \over 2}$

$\therefore$ $\angle AOM = {\tan ^{ - 1}}\left( {{{AM} \over {OM}}} \right)$

$ = {\tan ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right) = 30^\circ $

Electric flux passing from the whole cylinder

${\phi _1} = {{{q_{in}}} \over {{\varepsilon _0}}} = {{\lambda L} \over {{\varepsilon _0}}}$

$\therefore$ Electric flux passing through ABCD plane surface (shown only AB) = Electric flux passing through cylindrical surface ANB

$ = \left( {{{60^\circ } \over {360^\circ }}} \right)({\phi _1}) = {{\lambda L} \over {6{\varepsilon _0}}}$

$\therefore$ n = 6

Electric field at point P due to long uniformly charged solid cylinder is

${E_1} = {{\rho {R^2}} \over {2{\varepsilon _0}(2R)}} = {{\rho R} \over {4{\varepsilon _0}}}$

Electric field at point P due to spherical cavity is

${E_1} = {1 \over {4\pi {\varepsilon _0}}}{{\rho {4 \over 3}\pi {{\left( {{R \over 2}} \right)}^3}} \over {{{(2R)}^2}}} = {{\rho R} \over {96{\varepsilon _0}}}$

The electric field at the point P is

$ = {E_1} - {E_2}$

$ = {{\rho R} \over {4{\varepsilon _0}}} - {{\rho R} \over {96{\varepsilon _0}}} = {{\rho R} \over {4{\varepsilon _0}}}\left[ {1 - {1 \over {24}}} \right] = {{23\rho R} \over {96{\varepsilon _0}}} = {{23\rho R} \over {(16)6{\varepsilon _0}}} = {{23\rho R} \over {16k{\varepsilon _0}}}$

$\therefore$ $k = 6$

The net force on one of the charges due to other charges is

$F = {{2k{q^2}} \over {{a^2}}} + {{k{q^2}} \over {2{a^2}}} = {5 \over 2}\left( {{{k{q^2}} \over {{a^2}}}} \right)$

where $k = {1 \over {4\pi \varepsilon }}$. Here, as shown in the figure, line AB divided the soap film into two equal parts. The free-body diagram of half part is also depicted in the figure here.

At equilibrium, the surface tension balances the force.

Therefore,

${F_{surface}} = 2\sqrt 2 a\gamma $

That is, $2\sqrt 2 a\gamma = {5 \over 2}\left( {{{k{q^2}} \over {{a^2}}}} \right)$

$ \Rightarrow {a^3} = {5 \over {4\sqrt 2 }}\left( {{{{q^2}} \over \gamma }} \right)$

Therefore,

a = Any constant $ \times {\left( {{{{q^2}} \over \gamma }} \right)^{1/3}}$

Hence, N = 3.

Applying Gauss's theorem, we get

$E(4\pi {r^2}) = {{{q_{encl}}} \over {{t_0}}} = {1 \over t}\int\limits_0^r {k{x^2}(4\pi {x^2})dx} $

Now, $E{r^2} = {k \over {{t_0}}}\int\limits_0^r {{x^{2 + a}}dx = {k \over {{t_0}}}\left( {{{{r^{3 + a}}} \over {3 + a}}} \right)} $

Therefore, $E = {k \over {{t_0}}}{{{r^{1 + a}}} \over {3 + a}}$

That is, $E \propto {r^{1 + a}}$

Now, $E\left( {{R \over 2}} \right) = {1 \over 8}E(R)$

Therefore, ${\left( {{R \over 2}} \right)^{1 + a}} = {1 \over 8}{(R)^{1 + a}} \Rightarrow 8 = {2^{1 + a}}$

where $1 + a = 3$ and hence $a = 2$.