Electrostatics

AC = BC

V_D = V_E

We have, W = Q (V_E – V_D)

$ \Rightarrow $ W = 0

$\overrightarrow E = - \overrightarrow \nabla V$

where $\overrightarrow \nabla = \widehat i{\partial \over {\partial x}} + \widehat j{\partial \over {\partial y}} + {\partial \over {\partial z}}$

$\overrightarrow E = - \left[ {\widehat i{{\partial V} \over {\partial x}} + \widehat j{{\partial V} \over {\partial y}} + \widehat k{{\partial V} \over {\partial z}}} \right]$

Here, V = 4x²
$ \therefore \overrightarrow E = - 8x\widehat i$

The electric field at point (1, 0, 2) is

${\overrightarrow E _{\left( {1,0,2} \right)}} = - 8\widehat iV{m^{ - 1}}$

So electric field is along the negative X-axis.

According to Gauss’s law

${\phi _E} = {{{Q_{enclosed}}} \over {{\varepsilon _0}}}$

If the radius of the Gaussian surface is doubled, the outward electric flux will remain the same. This is because electric flux depends only on the charge enclosed by the surface.

Distance of point A from the two +q charges = L.

Distance of point A from the two –q charges
$ = \sqrt {{L^2} + {{\left( {2L} \right)}^2}} = \sqrt 5 L$



$ \therefore $ ${V_A} = \left( {{{Kq} \over L} \times 2} \right) - \left( {{{Kq} \over {\sqrt 5 L}} \times 2} \right)$

$ = {{2Kq} \over L}\left[ {1 - {1 \over {\sqrt 5 }}} \right]$

$ = {1 \over {4\pi {\varepsilon _0}}}.{{2q} \over L}\left( {1 - {1 \over {\sqrt 5 }}} \right)$

Electric field inside a charged conductor is always zero.

According to Coulomb’s law, the force of repulsion between the two positive ions each of charge q, separated by a distance d is given by

$F = {1 \over {4\pi {\varepsilon _0}}}{{\left( q \right)\left( q \right)} \over {{d^2}}}$

$F = {{{q^2}} \over {4\pi {\varepsilon _0}{d^2}}}$

${q^2} = 4\pi {\varepsilon _0}F{d^2}$

$q = \sqrt {4\pi {\varepsilon _0}F{d^2}} $   ...(i)

Since, q = ne
where, n = number of electrons missing from each ion
e = magnitude of charge on electron

$ \therefore n = {q \over e}$

$n = {{\sqrt {4\pi {\varepsilon _0}F{d^2}} } \over e}$   (Using (i))

= $\sqrt {{{4\pi {\varepsilon _0}F{d^2}} \over {{e^2}}}} $

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}}$