Electrostatics

Electric flux, $\phi $ = EA cos $\theta $ , where $\theta $ = angle between E and normal to the surface.

Here, $\theta = {\pi \over 2}$

$ \Rightarrow \phi $ = 0

The electric potential at a point,
V = –x²y – xz³ + 4.

The field
$\overrightarrow E = - \overrightarrow \nabla V = - \left( {{{\partial V} \over {\partial x}}\widehat i + {{\partial V} \over {\partial y}}\widehat j + {{\partial V} \over {\partial z}}\widehat k} \right)$

$ \therefore \overrightarrow E = \widehat i\left( {2xy + {z^3}} \right) + \widehat j{x^2} + \widehat k\left( {3x{z^2}} \right)$

${V_A} = {1 \over {4\pi {\varepsilon _0}}}\left\{ {{{qA} \over a} + {{qB} \over b} + {{qC} \over c}} \right\}$

$ = {{4\pi } \over {4\pi {\varepsilon _0}}}\left\{ {{{{a^2}\sigma } \over a} - {{{b^2}\sigma } \over b} + {{{c^2}\sigma } \over c}} \right\}$

${V_A} = {1 \over {{\varepsilon _0}}}\left\{ {{{{a^2}\sigma } \over a} - {{{b^2}\sigma } \over b} + {{{c^2}\sigma } \over c}} \right\}$

${V_B} = {1 \over {{\varepsilon _0}}}\left\{ {{{{a^2}\sigma } \over a} - {{{b^2}\sigma } \over b} + {{{c^2}\sigma } \over c}} \right\}$

${V_C} = {1 \over {{\varepsilon _0}}}\left\{ {{{{a^2}\sigma } \over a} - {{{b^2}\sigma } \over b} + {{{c^2}\sigma } \over c}} \right\}$

Given c = a + b.
If a = a, b = 2a and c = 3a for example, as c > b > a,

${V_A} = {1 \over {{\varepsilon _0}}}\left\{ {{{{a^2}\sigma } \over a} - {{4{b^2}\sigma } \over {2a}} + {{{c^2}\sigma } \over c}} \right\}$

${V_B} = {1 \over {{\varepsilon _0}}}\left\{ {{{{a^2}\sigma } \over {2a}} - {{4{a^2}\sigma } \over {2a}} + {{{c^2}\sigma } \over c}} \right\}$

${V_C} = {1 \over {{\varepsilon _0}}}\left\{ {{{{a^2}\sigma } \over {3a}} - {{4{a^2}\sigma } \over {3a}} + {{{c^2}\sigma } \over c}} \right\}$

It can seen by taking out common factors that
V_A = V_C > V_B i.e., V_A = V_C $ \ne $ V_B

By the symmetry of the figure, the electric fields at O due to the portions AC and BD are equal in magnitude and opposite in direction. So, they cancel each other.
Similarly, the field at O due to CD and AKB are equal in magnitude but opposite in direction. Therefore, the electric field at the centre due to the charge on the part ACDB is E along OK.

$V = {1 \over {4\pi {\varepsilon _0}}} \times {Q \over R}$

= Q × 10¹¹ volt    …(i)

$E = {1 \over {4\pi {\varepsilon _0}}} \times {Q \over {{R^2}}}$

= V/R = Q × 10¹¹ × ${4\pi {\varepsilon _0}}$ × 10¹¹    from ...(i)

= ${4\pi {\varepsilon _0}}$ × Q × 10²² volt/m

Potential at C = V_C = 0
Potential at D = V_D

$ = k\left( {{{ - q} \over L}} \right) + {{kq} \over {3L}} = - {2 \over 3}{{kq} \over L}$

Potential difference
V_D – V_C $ = - {2 \over 3}{{kq} \over L} = {1 \over {4\pi {\varepsilon _0}}}\left( { - {2 \over 3}.{q \over L}} \right)$

$ \Rightarrow $ Work done = Q (V_D – V_C)

$ = - {2 \over 3} \times {1 \over {4\pi {\varepsilon _0}}}{{qQ} \over L} = {{ - qQ} \over {6\pi {\varepsilon _0}L}}$

Let $\phi _A$, $\phi _B$ and $\phi _C$ are the electric flux linked with A, B and C.

According to Gauss theorem,

$\phi _A + \phi _B + \phi _C$ = ${q \over {{\varepsilon _0}}}$

Since $\phi _A$ = $\phi _C$

$ \Rightarrow 2{\phi _A} = {q \over {{\varepsilon _0}}} - {\phi _B}$

$ \Rightarrow 2{\phi _A} = {q \over {{\varepsilon _0}}} - \phi $   $\left( {Given\,{\phi _B} = \phi } \right)$

$ \therefore $ ${\phi _A} = {1 \over 2}\left( {{q \over {{\varepsilon _0}}} - \phi } \right)$

This consists of two dipoles, –q and +q with dipole moment along with the +y-direction and –q and +q along the x-direction.


$ \therefore $ The resultant moment = $\sqrt {{q^2}{a^2} + {q^2}{a^2}} = \sqrt 2 qa$

Along the direction 45° that is along OP where P is (+a, +a, 0).

Electric flux, ${\phi _E} = \int {\overrightarrow E .d\overrightarrow S } $

$ = \int {EdS\cos \theta } = \int {EdS\cos {{90}^o} = 0} $

The lines are parallel to the surface.

Work done in deflecting a dipole through an angle $\theta $ is given by

$W = \int\limits_0^\theta {pE\sin \theta d\theta } = pE\left( {1 - \cos \theta } \right)$

Since $\theta $ = 90°

$ \therefore $ W = pE(1 – cos90°) $ \Rightarrow $ W = pE

Work done is equal to zero because the
potential of A and B are the same = ${1 \over {4\pi {\varepsilon _0}}}{q \over a}$


No work is done if a particle does not change its potential energy.
i.e. initial potential energy = final potential energy.

We know that potential energy of discrete system of charges is given by

$U = {1 \over {4\pi {\varepsilon _0}}}\left( {{{{q_1}{q_2}} \over {{r_{12}}}} + {{{q_2}{q_3}} \over {{r_{23}}}} + {{{q_3}{q_1}} \over {{r_{31}}}}} \right)$

According to question,

${U_{initial}} = {1 \over {4\pi {\varepsilon _0}}}\left( {{{{q_1}{q_2}} \over {0.3}} + {{{q_2}{q_3}} \over {0.5}} + {{{q_3}{q_1}} \over {0.4}}} \right)$

${U_{final}} = {1 \over {4\pi {\varepsilon _0}}}\left( {{{{q_1}{q_2}} \over {0.3}} + {{{q_2}{q_3}} \over {0.1}} + {{{q_3}{q_1}} \over {0.4}}} \right)$

${U_{final}} - {U_{initial}} = {1 \over {4\pi {\varepsilon _0}}}\left( {{{{q_1}{q_2}} \over {0.1}} - {{{q_2}{q_3}} \over {0.5}}} \right)$

$ = {1 \over {4\pi {\varepsilon _0}}}\left[ {10{q_2}{q_3} - 2{q_2}{q_3}} \right] = {{{q_3}} \over {4\pi {\varepsilon _0}}}\left( {8{q_2}} \right)$

Using ${1 \over 2}m{v^2} = qV$

$V = {1 \over 2} \times {{2 \times {{10}^{ - 3}} \times 10 \times 10} \over {2 \times {{10}^{ - 6}}}} = 50\,kV$

When the dipole is in the direction of field then net force is qE + (–qE) = 0


and its potential energy is minimum = – P.E. = –qaE

The total flux through the cube ${\phi _{total}} = {q \over {{\varepsilon _0}}}$

$ \therefore $ The electric flux through any face
${\phi _{face}} = {q \over {6{\varepsilon _0}}} = {{4\pi q} \over {6\left( {4\pi {\varepsilon _0}} \right)}}$

There are eight corners of a cube and in each corner there is a charge of (–q). At the centre of the corner there is a charge of (+q). Each corner is equidistant from the centres of the cube and the distance (d) is half of the diagonals of the cube.

Diagonal of the cube = $\sqrt {{b^2} + {b^2} + {b^2}} = \sqrt 3 b$

$ \therefore $ $d = \sqrt 3 b/2$

Now, electric potential energy of the charge (+q) due to a charge (–q) at one corner

U $ = {{{q_1}{q_2}} \over {4\pi {\varepsilon _0}r}} = {{\left( { + q} \right) \times \left( { - q} \right)} \over {4\pi {\varepsilon _0}\left( {\sqrt 3 b/2} \right)}} = - {{{q^2}} \over {2\pi {\varepsilon _0}\left( {\sqrt 3 b} \right)}}$

$ \therefore $ Total electric potential energy due to all the eight identical charges
= $8U = - {{8{q^2}} \over {2\pi {\varepsilon _0}\sqrt 3 b}} = {{ - 4{q^2}} \over {\sqrt 3 \pi {\varepsilon _0}b}}$

Electric field intensity E is zero within a conductor due to charge given to it.

Also, $E = - {{dV} \over {dx}} \Rightarrow {{dV} \over {dx}} = 0$   (inside the conductor)

$ \therefore $ V = constant. [V is potential]

So potential remains same throughout the conductor.

For complete cube $\phi = {Q \over {{\varepsilon _0}}} \times {10^{ - 6}}$

For each face $\phi = {1 \over 6}{Q \over {{\varepsilon _0}}} \times {10^{ - 6}}$

When an electric dipole is placed in a uniform electrical field $\overrightarrow E $, the torque on the dipole is given by

$\overrightarrow \tau = \overrightarrow p \times \overrightarrow E $

According to Gauss's theorem, electric flux through a closed surface = ${Q \over {{\varepsilon _0}}}$ where q is charge enclosed by the surface.

Considering symmetric elements each of length dl at A and B, we note that electric fields perpendicular to PO are cancelled and those along PO are added. The electric field due to an element of length $dl( = ad\theta)$ along PO.

$dE = {1 \over {4\pi {\varepsilon _0}}}{{dq} \over {{a^2}}}\cos \theta $
$\left(\because {dl = ad\theta } \right)$

$ = {1 \over {4\pi {\varepsilon _0}}}{{\lambda dl} \over {{a^2}}}\cos \theta $

$ = {1 \over {4\pi {\varepsilon _0}}}{{\lambda \left( {ad\theta } \right)} \over {{a^2}}}\cos \theta $ Net electric field at O

$E = \int_{ - \pi /2}^{\pi /2} {dE} $

$ = 2\int_O^{\pi /2} {{1 \over {4\pi {\varepsilon _0}}}{{\lambda \left( {ad\theta } \right)} \over {{a^2}}}\cos \theta } $

$ = 2.{1 \over {4\pi {\varepsilon _0}}}{\lambda \over a}\left[ {\sin \theta } \right]_o^{\pi /2}$

$ = 2.{1 \over {4\pi {\varepsilon _0}}}{\lambda \over a}.1 = {\lambda \over {2\pi {\varepsilon _0}a}}$

As, $R$ be the radius of the ring. Consider a small strip of length $d l$ having charge $d q$ lying at an angle $\theta$.

$d l=R d \theta$

Charge on $d l=\lambda R d \theta$

Force at $1 \mathrm{~C}$ due to $\mathrm{dl}$

$=\frac{k \lambda R d \theta}{R^2}=\frac{k \lambda}{R} d \theta=d F$

We need to consider only the component $d F \cos \theta$, as the component $d F \sin \theta$ will cancel out because of the symmetrical element $d l$.

The total force on $1 \mathrm{C}$ is

$\begin{aligned} F & =\int_\limits{-\pi / 2}^{\pi / 2} d F \cos \theta \\ & =\frac{k \lambda}{R} \int_\limits{-\pi / 2}^{\pi / 2} \cos \theta d \theta \\ & =\frac{k \lambda}{R} \times 2=\frac{2 k \lambda}{R} \end{aligned}$

When the negative charge is shifted at a distance $x$ from the centre of the ring along its axis, then force acting on the point charge due to the ring.

$\begin{aligned} F & =q E ~(\text {towards centre}) \\\\ & =q \cdot \frac{k Q x}{\left(R^2+x^2\right)^{3 / 2}} \end{aligned}$

If $R >> x$, then $R^2+x^2 \simeq R^2$

and $\quad F=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q q x}{R^3}$ (towards centre)

$\Rightarrow \quad a=\frac{F}{m}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q q x}{m R^3}$

Since, restoring force $F_E \propto x$, therefore motion of charge particle will be SHM.

Time period of SHM,

$T=\frac{2 \pi}{\omega}=\left[\frac{16 \pi^3 \varepsilon_0 R^3 m}{Q q}\right]^{1 / 2}\left[\because \omega^2=\frac{Q q}{4 \pi \varepsilon_0 R^3}\right]$

Negatively charged body contains more electron than its natural state of mass is slightly increased.

In Ist case, when charge +Q is situated at C

Electric potential energy of system

$U_1=\frac{1}{4 \pi \varepsilon_0} \frac{q(-q)}{2 L}+\frac{1}{4 \pi \varepsilon_0} \frac{(-q) Q}{L}+\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{L}$

In Ind case, when charge $+Q$ is moved from $C$ to $D$

Electric potential energy of system in this case

$U_2=\frac{1}{4 \pi \varepsilon_0} \frac{q(-q)}{2 L}+\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{3 L}+\frac{1}{4 \pi \varepsilon_0} \frac{(-q)(Q)}{L}$

Work done $(\Delta U)=U_2-U_1$

$\begin{aligned} = & \left(-\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{2 L}+\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{3 L}-\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{L}\right) \\ & -\left(-\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{2 L}-\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{L}+\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{L}\right) \\ = & \frac{q Q}{4 \pi \varepsilon_0}\left[\frac{1}{3 L}-\frac{1}{L}\right]=\frac{q Q}{4 \pi \varepsilon_0} \frac{(1-3)}{3 L} \\ = & -\frac{2 q Q}{12 \pi \varepsilon_0 L}=-\frac{q Q}{6 \pi \varepsilon_0 L} \end{aligned}$

We know that, for an electric dipole

$\begin{aligned} E_{\text {axial }} & =\frac{1}{4 \pi \varepsilon_0}\left(\frac{2 p}{r^3}\right) \quad \text{..... (i)}\\ \text{and} \quad E_{\text {equatorial }} & =\frac{1}{4 \pi \varepsilon_0}\left(\frac{p}{r^3}\right) \quad \text{.... (ii)} \end{aligned}$

Hence, from Eqs. (i) and (ii), we get

$\frac{E_{\text {axil }}}{2}=E_{\text {equatorial }}$

Hence, $\quad E_{\text {equatorial }}=E / 2$

Reason is false as electric field due to dipole varies inversely as cube of distance, i.e.

$E \propto \frac{1}{r^3}$