Electric Dipole and Electric Field Questions in English

(B) The electric potential $V$ at a point on the axial line of an electric dipole at a distance $r$ from its center is given by the formula:
$V = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{p}{r^2}$
Where $p$ is the dipole moment.
From this expression,it is clear that the electric potential $V$ is inversely proportional to the square of the distance $r$.
Therefore,$V \propto \frac{1}{r^2}$ or $V \propto r^{-2}$.

(D) The electric field intensity on the axis of a dipole in air is given by $E_{\text{axis}} = \frac{2kP}{r^3}$.
Given $E_{\text{axis}} = 4 \text{ NC}^{-1}$,we have $4 = \frac{2kP}{r^3}$,which implies $\frac{kP}{r^3} = 2$ (Equation $i$).
The electric field intensity on the equatorial line of a dipole in a medium with dielectric constant $K$ is given by $E_{\text{eq}} = \frac{1}{K} \frac{kP}{r_1^3}$.
Here,$r_1 = 2r$ and $K = 4$.
Substituting these values,we get $E_{\text{eq}} = \frac{1}{4} \cdot \frac{kP}{(2r)^3} = \frac{1}{4} \cdot \frac{kP}{8r^3} = \frac{1}{32} \cdot \frac{kP}{r^3}$.
Using Equation $(i)$,$\frac{kP}{r^3} = 2$,so $E_{\text{eq}} = \frac{1}{32} \times 2 = \frac{1}{16} \text{ NC}^{-1}$.
Thus,the correct option is $(d)$.

(A) The dipole moment is $p = q \times d = (1 \times 10^{-6} \ C) \times (1 \ m) = 10^{-6} \ Cm$.
The moment of inertia $I$ of the system about the center is $I = m(d/2)^2 + m(d/2)^2 = m(d^2/2) = 1 \times (1^2/2) = 0.5 \ kg \ m^2$.
For a small angular displacement $\theta$, the restoring torque is $\tau = -pE \sin \theta \approx -pE \theta$.
Since $\tau = I \alpha$, we have $I \alpha = -pE \theta$, which gives $\alpha = -(pE/I) \theta$.
This represents simple harmonic motion with angular frequency $\omega = \sqrt{pE/I}$.
Substituting the values: $\omega = \sqrt{(10^{-6} \times 2 \times 10^4) / 0.5} = \sqrt{2 \times 10^{-2} / 0.5} = \sqrt{0.04} = 0.2 \ rad/s$.
The time taken to return to the mean position from the extreme position is $t = T/4 = (2 \pi / \omega) / 4 = \pi / (2 \omega)$.
$t = \pi / (2 \times 0.2) = \pi / 0.4 = 2.5 \pi \ s$.

(A) The electric field on the axial line of a short dipole at a distance $r$ is given by:
$E_{axial} = \frac{2kp}{r^3}$
The electric field on the equatorial line of a short dipole at a distance $2r$ is given by:
$E_{equatorial} = \frac{kp}{(2r)^3} = \frac{kp}{8r^3}$
According to the problem,$E_{axial} = x \cdot E_{equatorial}$.
Substituting the expressions:
$\frac{2kp}{r^3} = x \cdot \frac{kp}{8r^3}$
Canceling $kp/r^3$ from both sides:
$2 = \frac{x}{8}$
$x = 16$

(A) The arrangement of the electric dipole in the electric field will undergo Simple Harmonic Motion $(SHM)$ when the rod is set free. The time period $T$ of this $SHM$ is given by:
$T = 2\pi \sqrt{\frac{I}{pE}}$
where $I$ is the moment of inertia of the dipole,$p$ is the electric dipole moment,and $E$ is the electric field.
The moment of inertia $I$ about the center of the rod is $I = m(l/2)^2 + m(l/2)^2 = 2m(l/2)^2 = \frac{ml^2}{2}$.
Given: $m = 9.8 \times 10^{-3} \text{ kg}$,$l = 0.5 \text{ m}$,$q = 20 \times 10^{-6} \text{ C}$,$E = 12.1 \text{ N/C}$.
$I = \frac{9.8 \times 10^{-3} \times (0.5)^2}{2} = 4.9 \times 10^{-3} \times 0.25 = 1.225 \times 10^{-3} \text{ kg m}^2$.
$p = q \times l = 20 \times 10^{-6} \times 0.5 = 10^{-5} \text{ C m}$.
$T = 2\pi \sqrt{\frac{1.225 \times 10^{-3}}{10^{-5} \times 12.1}} = 2\pi \sqrt{\frac{1.225 \times 100}{12.1}} = 2\pi \sqrt{\frac{122.5}{12.1}} \approx 2\pi \sqrt{10.12} \approx 2\pi \times 3.18 \approx 20 \text{ s}$.
The time taken by the rod to move from the initial small angle to the parallel position (equilibrium position) is $T/4$.
Required time $= \frac{20}{4} = 5 \text{ s}$.

(D) The charges are $q_1 = 10 \mu C$ and $q_2 = -10 \mu C$. The distance between them is $2a = 10 \ cm$,so $a = 5 \ cm = 0.05 \ m$.
The point $P$ is on the perpendicular bisector at a distance $r = 12 \ cm = 0.12 \ m$ from the midpoint.
The distance from each charge to point $P$ is $d = \sqrt{r^2 + a^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \ cm = 0.13 \ m$.
The electric field due to each charge at $P$ has a magnitude $E = \frac{k|q|}{d^2} = \frac{9 \times 10^9 \times 10 \times 10^{-6}}{(0.13)^2} = \frac{9 \times 10^4}{0.0169} \approx 5.325 \times 10^6 \ N \ C^{-1}$.
The components of the electric fields perpendicular to the line $AB$ cancel out,while the components parallel to $AB$ add up.
The net electric field is $E_{\text{net}} = 2E \cos(\theta)$,where $\cos(\theta) = \frac{a}{d} = \frac{5}{13}$.
$E_{\text{net}} = 2 \times \left( \frac{9 \times 10^9 \times 10 \times 10^{-6}}{(0.13)^2} \right) \times \frac{5}{13} = 2 \times \frac{9 \times 10^4}{0.0169} \times \frac{5}{13} \approx 4.1 \times 10^6 \ N \ C^{-1}$.

(B) The line joining the two charges $+q$ and $-q$ forms an electric dipole.
Any point on the perpendicular bisector (equatorial line) of the dipole is equidistant from both charges.
Let the distance of a point $P$ on the perpendicular bisector from charge $+q$ be $r_1$ and from charge $-q$ be $r_2$. Since $P$ is on the perpendicular bisector, $r_1 = r_2 = r$.
The electric potential $V$ at point $P$ is given by $V = \frac{1}{4 \pi \varepsilon_0} (\frac{q}{r} + \frac{-q}{r}) = 0$.
However, the electric field $E$ at point $P$ is the vector sum of the fields due to $+q$ and $-q$. Since both fields are non-zero and directed such that they do not cancel out (the resultant field is parallel to the dipole axis), the electric field strength is non-zero.
Therefore, the electric potential is zero, but the electric field strength is non-zero.

(C) The charge $2q$ at vertex $B$ can be split into two charges of $+q$ each. One $+q$ charge forms a dipole with the $-q$ charge at vertex $A$,and the other $+q$ charge forms a dipole with the $-q$ charge at vertex $C$.
Let $AB = d_1$ and $BC = d_2$.
From the triangle,$d_1 = 10 \cos 30^{\circ} = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \text{ cm}$ and $d_2 = 10 \sin 30^{\circ} = 10 \times \frac{1}{2} = 5 \text{ cm}$.
The dipole moment $p_1$ (along $BA$) is $q \times d_1 = 5\sqrt{3}q \text{ C-cm}$.
The dipole moment $p_2$ (along $BC$) is $q \times d_2 = 5q \text{ C-cm}$.
Since these two dipoles are perpendicular to each other,the resultant dipole moment $P$ is:
$P = \sqrt{p_1^2 + p_2^2} = \sqrt{(5\sqrt{3}q)^2 + (5q)^2} = \sqrt{75q^2 + 25q^2} = \sqrt{100q^2} = 10q \text{ C-cm}$.

(C) The third statement is incorrect. An electric dipole in an external electric field experiences a torque that tends to align the dipole moment $p$ in the direction of the external electric field $E$. This alignment corresponds to the state of minimum potential energy $(U = -p \cdot E)$ and maximum stability. Therefore, the dipole does not orient itself in the opposite direction.

(B) The perpendicular bisector of the line joining two equal and opposite charges ($+q$ and $-q$) is known as the equatorial line of the electric dipole.
For any point on the equatorial line at a distance $r$ from the center of the dipole, the electric potential $V$ is given by the sum of potentials due to individual charges:
$V = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{r_1} + \frac{-q}{r_2} \right)$
Since any point on the perpendicular bisector is equidistant from both charges, $r_1 = r_2$, which implies $V = 0$.
However, the electric field strength $E$ at this point is not zero. It is directed parallel to the dipole axis (from $+q$ to $-q$) and is given by $E = \frac{1}{4 \pi \varepsilon_0} \frac{p}{r^3}$ (for $r \gg a$).

(B) The electric field $E$ due to an infinite charged cylinder at a distance $r$ from its axis is given by $E \propto \frac{1}{r}$.
Since the $+Q$ charge is closer to the cylinder than the $-Q$ charge,the electric field at the position of $+Q$ $(E_1)$ is stronger than the electric field at the position of $-Q$ $(E_2)$,i.e.,$E_1 > E_2$.
The force on the $+Q$ charge is $F_1 = Q E_1$ (directed away from the cylinder,i.e.,to the right).
The force on the $-Q$ charge is $F_2 = Q E_2$ (directed towards the cylinder,i.e.,to the left).
Since $E_1 > E_2$,the net force $F_{\text{net}} = F_1 - F_2$ is directed to the right.
Regarding torque,the force $F_1$ acts at a greater distance from the center of the rod and is larger,while $F_2$ acts at a smaller distance. The combination of these forces creates a clockwise torque about the center of the rod.

(B, C) In a uniform electric field $\vec{E}$,the force on charge $+q$ is $\vec{F}_+ = q\vec{E}$ and the force on charge $-q$ is $\vec{F}_- = -q\vec{E}$.
The net force on the dipole is $\vec{F}_{net} = \vec{F}_+ + \vec{F}_- = q\vec{E} - q\vec{E} = 0$. Thus,the net force on the dipole always vanishes.
The torque on the dipole is given by $\vec{\tau} = \vec{p} \times \vec{E}$. Since the torque is defined as the cross product of the dipole moment and the electric field,it is a physical quantity and is independent of the choice of the coordinate system.
Therefore,statements $(b)$ and $(c)$ are correct.

(A) The electric field at point $P$ at distance $x$ due to the three charges is:
$E = \frac{1}{4 \pi \epsilon_0} \left[ \frac{q}{(x-a)^2} - \frac{2q}{x^2} + \frac{q}{(x+a)^2} \right]$
$E = \frac{q}{4 \pi \epsilon_0} \left[ \frac{(x+a)^2 - 2(x^2-a^2) + (x-a)^2}{x^2(x^2-a^2)} \right]$
$E = \frac{q}{4 \pi \epsilon_0} \left[ \frac{x^2 + 2ax + a^2 - 2x^2 + 2a^2 + x^2 - 2ax + a^2}{x^2(x^2-a^2)} \right]$
$E = \frac{q}{4 \pi \epsilon_0} \left[ \frac{4a^2}{x^2(x^2-a^2)} \right]$
Since $x \gg a$,we can approximate $x^2 - a^2 \approx x^2$:
$E = \frac{q}{4 \pi \epsilon_0} \left[ \frac{4a^2}{x^2(x^2)} \right] = \frac{4qa^2}{4 \pi \epsilon_0 x^4}$
Given $qa^2 = Q$,we have $E = \frac{4Q}{4 \pi \epsilon_0 x^4}$.
Comparing this with $E = \frac{\alpha Q}{4 \pi \epsilon_0 x^\beta}$,we get $\alpha = 4$ and $\beta = 4$.
Therefore,$\alpha = \beta$.

(D) The dipole moments are $P_1 = q_1 \times l_1 = 2 \times 10^{-6} \text{ C} \times 10^{-2} \text{ m} = 2 \times 10^{-8} \text{ Cm}$ and $P_2 = q_2 \times l_2 = 4 \times 10^{-6} \text{ C} \times 10^{-2} \text{ m} = 4 \times 10^{-8} \text{ Cm}$.
Point $P$ is at a distance $r = 40 \text{ cm} = 0.4 \text{ m}$ from the centre of each dipole.
For dipole $A$,point $P$ is on its axial line. The electric field is $\vec{E}_1 = \frac{2KP_1}{r^3} \hat{i} = \frac{2 \times (9 \times 10^9) \times (2 \times 10^{-8})}{(0.4)^3} \hat{i} = \frac{360}{(0.064)} \hat{i} = 5625 \hat{i} \text{ N/C}$.
For dipole $B$,point $P$ is on its equatorial line. The electric field is $\vec{E}_2 = -\frac{KP_2}{r^3} \hat{j} = -\frac{(9 \times 10^9) \times (4 \times 10^{-8})}{(0.4)^3} \hat{j} = -\frac{360}{(0.064)} \hat{j} = -5625 \hat{j} \text{ N/C}$.
The net electric field is $\vec{E}_{net} = \vec{E}_1 + \vec{E}_2 = 5625(\hat{i} - \hat{j}) \text{ N/C}$.
The magnitude is $|\vec{E}_{net}| = 5625 \sqrt{1^2 + (-1)^2} = 5625 \sqrt{2} \text{ N/C}$.
Since $5625 = \frac{9 \times 10^4}{16}$,the magnitude is $\frac{9}{16} \sqrt{2} \times 10^4 \text{ N/C}$.

(C) The dipole moment $\vec{p}$ of a system of charges is given by $\vec{p} = \sum q_i \vec{r}_i$.
Given charges are $q_1 = +2q$ at $\vec{r}_1 = (0, -3a) = -3a\hat{j}$,$q_2 = +3q$ at $\vec{r}_2 = (2a, 0) = 2a\hat{i}$,and $q_3 = -4q$ at $\vec{r}_3 = (-2a, 0) = -2a\hat{i}$.
Substituting these values into the formula:
$\vec{p} = (2q)(-3a\hat{j}) + (3q)(2a\hat{i}) + (-4q)(-2a\hat{i})$
$\vec{p} = -6qa\hat{j} + 6qa\hat{i} + 8qa\hat{i}$
$\vec{p} = (6qa + 8qa)\hat{i} - 6qa\hat{j}$
$\vec{p} = 14qa\hat{i} - 6qa\hat{j}$
$\vec{p} = 2qa(7\hat{i} - 3\hat{j})$

(C) The work done $W$ in rotating an electric dipole in a uniform electric field is given by the formula: $W = PE(\cos \theta_1 - \cos \theta_2)$.
Initially,the dipole is parallel to the electric field,so the initial angle $\theta_1 = 0^\circ$.
Finally,the dipole is rotated by an angle of $90^\circ$,so the final angle $\theta_2 = 90^\circ$.
Substituting these values into the formula:
$W = PE(\cos 0^\circ - \cos 90^\circ)$
Since $\cos 0^\circ = 1$ and $\cos 90^\circ = 0$,we get:
$W = PE(1 - 0) = PE$.

(A) The frequency of oscillation for a dipole in a uniform electric field $\vec{E}$ is given by $\nu = \frac{1}{2\pi}\sqrt{\frac{pE}{I}}$,where $p$ is the dipole moment and $I$ is the moment of inertia.
Initially,the electric field is $\vec{E}_1 = E_0\hat{i}$,so its magnitude is $E_1 = E_0$.
The frequency is $\nu_1 = \frac{1}{2\pi}\sqrt{\frac{pE_0}{I}}$.
When the second field $\vec{E}_2 = 2E_0\hat{j} + 2E_0\hat{k}$ is added,the resultant electric field is $\vec{E}_{res} = E_0\hat{i} + 2E_0\hat{j} + 2E_0\hat{k}$.
The magnitude of the resultant field is $E_{res} = \sqrt{E_0^2 + (2E_0)^2 + (2E_0)^2} = \sqrt{E_0^2 + 4E_0^2 + 4E_0^2} = \sqrt{9E_0^2} = 3E_0$.
The new frequency is $\nu_2 = \frac{1}{2\pi}\sqrt{\frac{p(3E_0)}{I}} = \sqrt{3} \nu_1$.
The percentage change in frequency is $\frac{\nu_2 - \nu_1}{\nu_1} \times 100 = (\sqrt{3} - 1) \times 100 \approx (1.732 - 1) \times 100 = 73.2\%$.
Thus,the approximate percentage change is $73\%$.

(B) For a short electric dipole,the electric field at a point at distance $r$ and angle $\theta$ from the dipole axis is given by $\vec{E} = E_r \hat{r} + E_\theta \hat{\theta}$,where $E_r = \frac{2kp\cos\theta}{r^3}$ and $E_\theta = \frac{kp\sin\theta}{r^3}$.
For dipole $A$,the point $x$ lies on its axial line,so $\theta = 0^{\circ}$. The electric field due to $A$ is $E_A = \frac{2kp_1}{r^3}$ directed along the line $Ox$.
For dipole $B$,the point $x$ lies on its equatorial line,so $\theta = 90^{\circ}$. The electric field due to $B$ is $E_B = \frac{kp_2}{r^3}$ directed perpendicular to the line $Ox$.
The resultant electric field makes an angle of $60^{\circ}$ with the line $Ox$. Therefore,$\tan 60^{\circ} = \frac{E_B}{E_A}$.
Substituting the values,we get $\sqrt{3} = \frac{kp_2/r^3}{2kp_1/r^3} = \frac{p_2}{2p_1}$.
Thus,the ratio $\frac{p_2}{p_1} = 2\sqrt{3}$.