Electric Dipole and Electric Field Questions in English

(C) The potential energy $U$ of an electric dipole in a uniform electric field $E$ is given by the formula $U = -p \cdot E = -pE \cos \theta$,where $p$ is the dipole moment and $\theta$ is the angle between the dipole moment vector and the electric field vector.
To minimize the potential energy $U$,the value of $\cos \theta$ must be maximum.
The maximum value of $\cos \theta$ is $1$,which occurs when $\theta = 0$.
Therefore,the potential energy is minimum when the dipole moment is aligned in the direction of the electric field,i.e.,$\theta = 0$.

(A) In a non-uniform electric field,the electric field intensity at the position of the two charges of the dipole ($-q$ and $+q$) is different.
Since $F = qE$,the forces acting on the two charges are $F_1 = -qE_1$ and $F_2 = qE_2$.
Because $E_1
eq E_2$,the net force $F_{net} = F_1 + F_2
eq 0$.
Additionally,since these forces act at different points and are not collinear,they exert a net torque on the dipole.
Therefore,the dipole experiences both a net force and a net torque.

(B) The electric dipole moment $p$ is given by $p = q \times (2a)$,where $q = 2 \times 10^{-6} \, C$ and the separation distance $2a = 3 \, cm = 3 \times 10^{-2} \, m$.
$p = (2 \times 10^{-6} \, C) \times (3 \times 10^{-2} \, m) = 6 \times 10^{-8} \, C \cdot m$.
The maximum torque $\tau_{max}$ on a dipole in an electric field $E$ is given by $\tau_{max} = pE$.
Given $E = 2 \times 10^5 \, N/C$.
$\tau_{max} = (6 \times 10^{-8} \, C \cdot m) \times (2 \times 10^5 \, N/C) = 12 \times 10^{-3} \, N \cdot m$.

(D) The work done $W$ in rotating an electric dipole in an external electric field is given by the formula:
$W = pE(\cos \theta_1 - \cos \theta_2)$
Here,the initial angle $\theta_1 = 90^\circ$ (since it is placed normal to the field).
The dipole is deflected through an angle of $180^\circ$,so the final angle $\theta_2 = 90^\circ + 180^\circ = 270^\circ$.
Substituting these values:
$W = pE(\cos 90^\circ - \cos 270^\circ)$
Since $\cos 90^\circ = 0$ and $\cos 270^\circ = 0$,
$W = pE(0 - 0) = 0$.
Therefore,the work done is zero.

(D) The electric field intensity $E$ on the axial line of a dipole at a distance $d$ from its center is given by the formula:
$E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{2p}{d^3}$
Since the dipole moment $p$ is defined as the product of the magnitude of one charge $q$ and the separation distance $r$ between them $(p = q \times r)$,we substitute this into the formula:
$E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{2(q \times r)}{d^3}$
From this expression,it is clear that the electric field intensity $E$ is proportional to $\frac{qr}{d^3}$.
Therefore,the correct option is $D$.

(B) The electric dipole moment $p$ is defined as the product of the magnitude of one of the charges $q$ and the distance of separation $2l$ between them.
Given: Charge of an electron or proton $q = 1.6 \times 10^{-19}\,C$.
Distance $2l = 1\,\mathring{A} = 10^{-10}\,m$.
Therefore,the dipole moment $p = q \times 2l = (1.6 \times 10^{-19}\,C) \times (10^{-10}\,m) = 1.6 \times 10^{-29}\,C \cdot m$.
Thus,the correct option is $B$.

(B) The electric field $E$ at an axial point of an electric dipole at a distance $r$ from its center is given by the formula:
$E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{2pr}{ (r^2 - a^2)^2 }$
For a short dipole where $r \gg a$,the expression simplifies to:
$E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{2p}{r^3}$
Thus,the electric field $E$ is inversely proportional to the cube of the distance $r$ (i.e.,$E \propto \frac{1}{r^3}$).
Therefore,the correct option is $B$.

(C) The electric dipole moment $p$ is defined as the product of the magnitude of one of the charges and the distance between them.
$p = q \times (2l)$
Given:
Charge $q = 3.2 \times 10^{-19} \text{ C}$
Separation distance $2l = 2.4 \text{ Å} = 2.4 \times 10^{-10} \text{ m}$
Calculating the dipole moment:
$p = (3.2 \times 10^{-19} \text{ C}) \times (2.4 \times 10^{-10} \text{ m})$
$p = 7.68 \times 10^{-29} \text{ C} \cdot \text{m}$

(C) For an electric dipole placed at the origin along the $x$-axis,the electric field components at a point $P(r, \theta)$ are given by:
Radial component: $E_r = \frac{2p \cos \theta}{4 \pi \epsilon_0 r^3}$
Transverse component: $E_{\theta} = \frac{p \sin \theta}{4 \pi \epsilon_0 r^3}$
The angle $\alpha$ that the resultant electric field makes with the position vector (radial direction) is given by $\tan \alpha = \frac{E_{\theta}}{E_r} = \frac{1}{2} \tan \theta$.
The position vector makes an angle $\theta$ with the $x$-axis.
The angle $\phi$ that the electric field makes with the $x$-axis is the sum of the angle of the position vector with the $x$-axis and the angle $\alpha$ between the electric field and the position vector.
Therefore,$\phi = \theta + \alpha$.

(B) When an electric dipole with dipole moment $\overrightarrow{P}$ is placed in an external uniform electric field $\overrightarrow{E}$,it experiences a torque $\overrightarrow{\tau}$.
By definition,the torque is the cross product of the dipole moment and the electric field.
The formula for torque is given by $\overrightarrow{\tau} = \overrightarrow{P} \times \overrightarrow{E}$.
Therefore,the correct option is $(b)$.

(B) The electric dipole moment $\vec{p}$ is directed from the negative charge to the positive charge.
For a point on the equatorial line of an electric dipole,the electric field $\vec{E}$ is given by $\vec{E} = -\frac{1}{4\pi\epsilon_0} \frac{\vec{p}}{r^3}$.
The negative sign indicates that the direction of the electric field at any point on the equatorial line is opposite to the direction of the dipole moment $\vec{p}$.
Thus,the electric field and the dipole moment are anti-parallel.

(D) Given: Charge $q = 4 \times 10^{-8} \, C$,separation $2a = 2 \times 10^{-2} \, cm = 2 \times 10^{-4} \, m$,and electric field $E = 4 \times 10^8 \, N/C$.
Dipole moment $p = q \times 2a = (4 \times 10^{-8}) \times (2 \times 10^{-4}) = 8 \times 10^{-12} \, Cm$.
Maximum torque $\tau_{max} = pE = (8 \times 10^{-12}) \times (4 \times 10^8) = 32 \times 10^{-4} \, Nm$.
Work done in rotating the dipole from $0^\circ$ to $180^\circ$ is $W = pE(1 - \cos 180^\circ) = pE(1 - (-1)) = 2pE$.
$W = 2 \times (32 \times 10^{-4}) = 64 \times 10^{-4} \, J$.

(B) The electric field strength of a short dipole at a point on its axial line at a distance $r$ is given by ${E_a} = \frac{1}{4\pi\epsilon_0} \cdot \frac{2p}{r^3}$.
The electric field strength of a short dipole at a point on its equatorial line at the same distance $r$ is given by ${E_e} = \frac{1}{4\pi\epsilon_0} \cdot \frac{p}{r^3}$.
Comparing the two expressions,we can see that ${E_a} = 2 \times \left( \frac{1}{4\pi\epsilon_0} \cdot \frac{p}{r^3} \right) = 2{E_e}$.
Therefore,the correct relationship is ${E_a} = 2{E_e}$.

(D) point charge $q$ produces a non-uniform electric field given by $E = \frac{kq}{r^2}$.
Since the electric field is non-uniform,the force on the two charges of the dipole ($+q$ and $-q$) will not be equal and opposite,so the net force on the dipole cannot be zero.
However,the torque on the dipole is given by $\vec{\tau} = \vec{p} \times \vec{E}$.
If the dipole is placed such that its dipole moment $\vec{p}$ is parallel or anti-parallel to the electric field $\vec{E}$ at that point,the torque $\vec{\tau}$ will be zero.
Therefore,the torque on the dipole due to the field may be zero.

(D) The electric field at a point on the equatorial line (perpendicular bisector) of an electric dipole is given by the formula:
$E_{equatorial} = \frac{1}{4\pi\epsilon_0} \frac{p}{r^3}$
where $p$ is the dipole moment and $r$ is the distance from the center of the dipole.
From this expression,it is clear that the electric field $E$ is directly proportional to the dipole moment $p$ $(E \propto p)$ and inversely proportional to the cube of the distance $r$ $(E \propto r^{-3})$.
Therefore,the electric field at $Q$ is proportional to $p$ and $r^{-3}$.

(D) The electric field intensity on the axial line of a dipole at a distance $x$ is given by $E_{axial} = \frac{1}{4\pi\epsilon_0} \cdot \frac{2p}{x^3}$.
The electric field intensity on the equatorial line of a dipole at a distance $y$ is given by $E_{equatorial} = \frac{1}{4\pi\epsilon_0} \cdot \frac{p}{y^3}$.
Given that the magnitudes are equal,we have $E_{axial} = E_{equatorial}$.
Substituting the expressions: $\frac{2p}{x^3} = \frac{p}{y^3}$.
Simplifying the equation: $\frac{2}{x^3} = \frac{1}{y^3}$.
Rearranging for the ratio: $\frac{x^3}{y^3} = 2$.
Taking the cube root on both sides: $\frac{x}{y} = \sqrt[3]{2}$.
Therefore,the ratio $x:y$ is $\sqrt[3]{2}:1$.

(C) In a uniform electric field,the force on the positive charge is $F_+ = qE$ and the force on the negative charge is $F_- = -qE$. The net force on the dipole is $F_{net} = F_+ + F_- = qE - qE = 0$. However,since these two forces act at different points,they form a couple that exerts a torque on the dipole given by $\tau = pE \sin \theta$. Therefore,the dipole experiences only a torque and no net force.

(A) Given: Dipole length $2l = 10 \, cm = 0.1 \, m$,so $l = 0.05 \, m$. Charge $q = 500 \, \mu C = 5 \times 10^{-4} \, C$. Distance from the center of the dipole $r = 20 \, cm + 5 \, cm = 25 \, cm = 0.25 \, m$.
Dipole moment $p = q \times 2l = (5 \times 10^{-4} \, C) \times (0.1 \, m) = 5 \times 10^{-5} \, C \cdot m$.
The electric field on the axial line of a dipole is given by $E = \frac{1}{4\pi\epsilon_0} \cdot \frac{2pr}{(r^2 - l^2)^2}$.
Substituting the values: $E = (9 \times 10^9) \times \frac{2 \times (5 \times 10^{-5}) \times 0.25}{((0.25)^2 - (0.05)^2)^2}$.
$E = (9 \times 10^9) \times \frac{2.5 \times 10^{-5}}{(0.0625 - 0.0025)^2} = (9 \times 10^9) \times \frac{2.5 \times 10^{-5}}{(0.06)^2}$.
$E = \frac{22.5 \times 10^4}{0.0036} = 6.25 \times 10^7 \, N/C$.

(D) The electric potential $V$ at a point due to an electric dipole is given by the formula: $V = \frac{kp \cos \theta}{r^2}$,where $p$ is the dipole moment,$r$ is the distance from the center of the dipole,and $\theta$ is the angle between the dipole axis and the position vector of the point.
For the potential to be maximum,$\cos \theta$ must be maximum $(+1)$,which occurs at $\theta = 0^\circ$.
For the potential to be minimum,$\cos \theta$ must be minimum $(-1)$,which occurs at $\theta = 180^\circ$.
Therefore,the potential is maximum at $0^\circ$ and minimum at $180^\circ$.

(C) In a uniform electric field,the force on a dipole is given by $\vec{F} = q\vec{E} + (-q)\vec{E} = 0$.
Since the dipole moment $\vec{p}$ is along the direction of the electric field $\vec{E}$,the angle $\theta$ between them is $0^\circ$.
The potential energy $U$ of an electric dipole in an external electric field is given by $U = -\vec{p} \cdot \vec{E} = -pE \cos \theta$.
Substituting $\theta = 0^\circ$,we get $U = -pE \cos(0^\circ) = -pE$.
Since $-pE$ is the lowest possible value for the potential energy,it is the minimum potential energy.
Therefore,the force is zero and the potential energy is minimum.

(B) The electric field intensity $E$ at a point due to an electric dipole is given by the formula:
$E = \frac{p}{4 \pi \varepsilon_{0} r^{3}} \sqrt{3 \cos^{2} \theta + 1}$
where $p$ is the dipole moment,$r$ is the distance from the center of the dipole,and $\theta$ is the angle with the dipole axis.
From this expression,it is clear that $E \propto \frac{1}{r^{3}}$.
Therefore,the intensity of the electric field is inversely proportional to the cube of the distance $r$.

(B) The electric field on the axis of an electric dipole at a distance $r$ from the center is given by $E_a = \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3}$.
The electric field on the equatorial plane of an electric dipole at a distance $r$ from the center is given by $E_e = \frac{1}{4\pi\epsilon_0} \frac{p}{r^3}$.
Taking the ratio of the two fields:
$\frac{E_a}{E_e} = \frac{\frac{1}{4\pi\epsilon_0} \frac{2p}{r^3}}{\frac{1}{4\pi\epsilon_0} \frac{p}{r^3}} = \frac{2}{1}$.
Therefore,the ratio is $2:1$.

(C) The dipole moment $p$ is given by $p = q \times d$.
Given $q = 2 \times 10^{-6} \, C$ and $d = 0.01 \, m$.
$p = (2 \times 10^{-6} \, C) \times (0.01 \, m) = 2 \times 10^{-8} \, C \cdot m$.
The torque $\tau$ on a dipole is given by $\tau = pE \sin \theta$.
For maximum torque,$\sin \theta = 1$,so $\tau_{\max} = pE$.
Substituting the values: $\tau_{\max} = (2 \times 10^{-8} \, C \cdot m) \times (5 \times 10^5 \, N/C)$.
$\tau_{\max} = 10 \times 10^{-3} \, N \cdot m$.

(D) The potential energy $U$ of a dipole in an electric field is given by $U = -pE \cos \theta$,where $\theta$ is the angle between the dipole moment $p$ and the electric field $E$.
Initially,the dipole is parallel to the field,so $\theta_1 = 0^\circ$.
The initial potential energy is $U_1 = -pE \cos(0^\circ) = -pE$.
Finally,the dipole is anti-parallel to the field,so $\theta_2 = 180^\circ$.
The final potential energy is $U_2 = -pE \cos(180^\circ) = -pE(-1) = pE$.
The work done by an external agent is equal to the change in potential energy: $W = U_2 - U_1$.
$W = pE - (-pE) = 2pE$.

(B) An electric dipole consists of two equal and opposite charges separated by a small distance.
Since the charges are stationary,they produce only an electrostatic field in the surrounding region.
According to the principles of electromagnetism,a magnetic field is produced only by moving charges (currents) or time-varying electric fields.
Therefore,a stationary electric dipole produces an electric field only.

(A) Let the neutral point $N$ be at a distance $x$ from the dipole of moment $P$ and at a distance $(25 - x)$ from the dipole of moment $64 P$.
At the neutral point $N$,the magnitude of the electric field due to dipole $1$ must be equal to the magnitude of the electric field due to dipole $2$.
The electric field on the axial line of a dipole is given by $E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{2p}{r^3}$.
Equating the fields:
$\frac{1}{4\pi \varepsilon_0} \cdot \frac{2P}{x^3} = \frac{1}{4\pi \varepsilon_0} \cdot \frac{2(64P)}{(25 - x)^3}$
$\frac{1}{x^3} = \frac{64}{(25 - x)^3}$
Taking the cube root on both sides:
$\frac{1}{x} = \frac{4}{25 - x}$
$25 - x = 4x$
$5x = 25$
$x = 5 \ cm$.

(A) The torque $\vec{\tau}$ acting on an electric dipole placed in a uniform electric field is given by the cross product: $\vec{\tau} = \vec{P} \times \vec{E}$.
The magnitude of the torque is given by $\tau = PE \sin \theta$,where $\theta$ is the angle between the dipole moment vector $\vec{P}$ and the electric field vector $\vec{E}$.
For the torque to be maximum,$\sin \theta$ must be maximum.
The maximum value of $\sin \theta$ is $1$,which occurs when $\theta = 90^{\circ}$.
Therefore,the torque is maximum when the angle between $\vec{P}$ and $\vec{E}$ is $90^{\circ}$.

(C) The electric dipole moment $\vec{p}$ is directed from the negative charge to the positive charge.
At any point on the equatorial line of an electric dipole,the direction of the electric field $\vec{E}$ is parallel to the dipole axis but directed from the positive charge towards the negative charge.
Therefore,the direction of the electric field intensity at a point on the equatorial line is exactly opposite to the direction of the dipole moment.
Since the vectors are anti-parallel,the angle between them is $180^o$.

(C) The electric field due to a dipole at an axial point at distance $r$ is given by $E_{\text{axial}} = \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3} = E$.
When the dipole is rotated by $90^{\circ}$ about its perpendicular axis,the point that was previously on the axial line now lies on the equatorial line of the dipole.
The electric field due to a dipole at an equatorial point at distance $r$ is given by $E_{\text{equatorial}} = \frac{1}{4\pi\epsilon_0} \frac{p}{r^3}$.
Comparing the two expressions,we get $E_{\text{equatorial}} = \frac{E_{\text{axial}}}{2} = \frac{E}{2}$.

(A) When an electric dipole is placed in a uniform electric field $E$,it experiences a torque $\tau = pE \sin \theta$.
For a small angular displacement $\theta$ from the equilibrium position,$\sin \theta \approx \theta$.
Thus,the restoring torque is $\tau = -pE\theta$.
Using Newton's second law for rotation,$\tau = I \alpha = I \frac{d^2\theta}{dt^2}$.
Equating the two expressions: $I \frac{d^2\theta}{dt^2} = -pE\theta$.
Rearranging gives $\frac{d^2\theta}{dt^2} = -(\frac{pE}{I})\theta$.
Comparing this with the standard simple harmonic motion equation $\frac{d^2\theta}{dt^2} = -\omega^2 \theta$,we get $\omega^2 = \frac{pE}{I}$.
Therefore,the angular frequency is $\omega = \sqrt{\frac{pE}{I}}$.

(B) $1$. In a uniform electric field $E$, the force on the positive charge is $F_+ = qE$ and the force on the negative charge is $F_- = -qE$.
$2$. The net force on the dipole is $F_{net} = F_+ + F_- = qE - qE = 0$.
$3$. The potential energy $U$ of a dipole in an electric field is given by $U = -p \cdot E = -pE \cos \theta$.
$4$. When the dipole moment $p$ is aligned with the electric field $E$, the angle $\theta = 0^\circ$.
$5$. Thus, $U = -pE \cos(0^\circ) = -pE$, which is the minimum possible potential energy.

(A) For an electric dipole with dipole moment $\vec{p}$,the electric field at a point on its axial line at a distance $r$ from the center (where $r \gg a$) is given by the formula:
$\vec{E}_{axial} = \frac{1}{4\pi\epsilon_0} \frac{2\vec{p}}{r^3}$.
This field is directed along the direction of the dipole moment $\vec{p}$.

(B) Consider two charges $+q$ and $-q$ separated by a distance $2a$. Let $P$ be a point on the perpendicular bisector at a distance $x$ from the midpoint.
The electric potential $V$ at point $P$ due to a charge $q$ at distance $r$ is given by $V = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r}$.
The distance of point $P$ from both charges $+q$ and $-q$ is $r = \sqrt{x^2 + a^2}$.
The potential due to $+q$ is $V_1 = \frac{1}{4 \pi \varepsilon_0} \frac{q}{\sqrt{x^2 + a^2}}$.
The potential due to $-q$ is $V_2 = -\frac{1}{4 \pi \varepsilon_0} \frac{q}{\sqrt{x^2 + a^2}}$.
The total electric potential at point $P$ is $V = V_1 + V_2 = \frac{1}{4 \pi \varepsilon_0} \frac{q}{\sqrt{x^2 + a^2}} - \frac{1}{4 \pi \varepsilon_0} \frac{q}{\sqrt{x^2 + a^2}} = 0$.
Thus,the electric potential at any point on the perpendicular bisector of an electric dipole is zero.

(A) The electric field exerts a torque on the dipole given by $\tau = pE \sin \theta$.
The work done $W$ in rotating the dipole from an initial angle $\theta_1 = 0^\circ$ (equilibrium position) to a final angle $\theta_2 = \theta$ is given by the integral of the torque with respect to the angle:
$W = \int_{0}^{\theta} \tau \, d\theta$
$W = \int_{0}^{\theta} pE \sin \theta \, d\theta$
$W = pE [-\cos \theta]_{0}^{\theta}$
$W = -pE (\cos \theta - \cos 0^\circ)$
Since $\cos 0^\circ = 1$,we get:
$W = -pE (\cos \theta - 1) = pE(1 - \cos \theta)$.

(B) An electric dipole consists of two equal and opposite charges,$+q$ and $-q$,separated by a distance $2a$.
Let the midpoint of the dipole be the origin $O$.
The electric potential $V$ at any point due to a dipole is given by $V = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2}$.
At the midpoint of the axis,the distance from the positive charge is $a$ and from the negative charge is $a$.
The potential due to $+q$ is $V_+ = \frac{1}{4\pi\epsilon_0} \frac{q}{a}$ and the potential due to $-q$ is $V_- = -\frac{1}{4\pi\epsilon_0} \frac{q}{a}$.
The total potential $V = V_+ + V_- = \frac{1}{4\pi\epsilon_0} \frac{q}{a} - \frac{1}{4\pi\epsilon_0} \frac{q}{a} = 0$.
Thus,the electric potential at the midpoint of an electric dipole is zero.

(A) The torque $\vec{\tau}$ acting on an electric dipole placed in a uniform electric field is given by the cross product: $\vec{\tau} = \vec{p} \times \vec{E}$.
The magnitude of the torque is given by: $\tau = pE \sin(\theta)$,where $\theta$ is the angle between the dipole moment $\vec{p}$ and the electric field $\vec{E}$.
For the torque to be maximum,$\sin(\theta)$ must be maximum.
The maximum value of $\sin(\theta)$ is $1$,which occurs when $\theta = 90^{\circ}$.
Therefore,the torque is maximum when the angle between $\vec{p}$ and $\vec{E}$ is $90^{\circ}$.

(C) The electric potential due to a dipole at a point $(r, \theta)$ is given by $V = \frac{1}{4\pi \varepsilon_0} \frac{p \cos \theta}{r^2}$.
Potential at point $A$ $(r, 135^{\circ})$:
$V_A = \frac{1}{4\pi \varepsilon_0} \frac{p \cos 135^{\circ}}{r^2} = \frac{1}{4\pi \varepsilon_0} \frac{p (-1/\sqrt{2})}{r^2}$.
Potential at point $B$ $(r, 45^{\circ})$:
$V_B = \frac{1}{4\pi \varepsilon_0} \frac{p \cos 45^{\circ}}{r^2} = \frac{1}{4\pi \varepsilon_0} \frac{p (1/\sqrt{2})}{r^2}$.
The work done by an external agent in moving a charge $q$ from $A$ to $B$ is $W = q(V_B - V_A)$.
$W = q \left[ \frac{1}{4\pi \varepsilon_0} \frac{p}{r^2} \left( \frac{1}{\sqrt{2}} - \left( -\frac{1}{\sqrt{2}} \right) \right) \right]$
$W = q \left[ \frac{1}{4\pi \varepsilon_0} \frac{p}{r^2} \left( \frac{2}{\sqrt{2}} \right) \right]$
$W = \frac{1}{4\pi \varepsilon_0} \frac{\sqrt{2} qp}{r^2}$.

(D) In a uniform electric field,an electric dipole experiences only a torque,and the net force is zero.
However,in a non-uniform electric field,the electric field strength varies at different points.
Since the two charges of the dipole ($+q$ and $-q$) are at different positions,they experience different electric forces $(F = qE)$.
Therefore,the net force on the dipole is non-zero.
Additionally,because the forces are not collinear,the dipole also experiences a torque.
Thus,in a non-uniform electric field,an electric dipole experiences both a net force and a torque.

(B) The electric dipole moment $(p)$ is defined as the product of the magnitude of one of the charges $(q)$ and the distance of separation $(2a)$ between them.
Mathematically,$p = q \times 2a$.
The $SI$ unit of charge $(q)$ is Coulomb $(C)$ and the $SI$ unit of distance $(2a)$ is meter $(m)$.
Therefore,the $SI$ unit of electric dipole moment is $C \cdot m$ (Coulomb-meter).

(B) The electric field due to a short electric dipole at a distance $r$ on its axial line is given by $E_{\text{axial}} = \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3}$.
The electric field due to a short electric dipole at a distance $r$ on its equatorial line is given by $E_{\text{equatorial}} = \frac{1}{4\pi\epsilon_0} \frac{p}{r^3}$.
Taking the ratio of the axial field to the equatorial field:
$\frac{E_{\text{axial}}}{E_{\text{equatorial}}} = \frac{\frac{1}{4\pi\epsilon_0} \frac{2p}{r^3}}{\frac{1}{4\pi\epsilon_0} \frac{p}{r^3}} = \frac{2}{1}$.
Therefore,the ratio is $2 : 1$.

(D) The electric field $E$ produced by a point charge is non-uniform,as it varies with distance $r$ $(E \propto 1/r^2)$.
The force on a dipole in a non-uniform electric field is given by $\vec{F} = (\vec{p} \cdot \nabla) \vec{E}$. Since the field is non-uniform,the net force on the dipole is generally non-zero.
The torque on a dipole is given by $\vec{\tau} = \vec{p} \times \vec{E}$.
If the dipole moment $\vec{p}$ is parallel or anti-parallel to the electric field $\vec{E}$,the angle $\theta$ between them is $0^\circ$ or $180^\circ$. Since $\tau = pE \sin \theta$,in these specific orientations,$\sin(0^\circ) = 0$ or $\sin(180^\circ) = 0$,making the torque zero.
Therefore,the torque on the dipole may be zero.

(B) The electric field $\vec{E}$ at a point on the equatorial line of an electric dipole is given by the formula $\vec{E} = -\frac{1}{4\pi\epsilon_0} \frac{\vec{p}}{r^3}$,where $\vec{p}$ is the electric dipole moment.
The negative sign indicates that the direction of the electric field $\vec{E}$ is opposite to the direction of the dipole moment $\vec{p}$.
Therefore,the electric field is anti-parallel to the dipole moment.

(B) The electric potential $V$ due to a short electric dipole at a point $(r, \theta)$ is given by the formula:
$V = \frac{kp \cos \theta}{r^2}$
where $k = \frac{1}{4\pi \epsilon_0}$ is the Coulomb constant,$p$ is the dipole moment,$r$ is the distance from the center of the dipole,and $\theta$ is the angle between the position vector of the point and the dipole axis.
Thus,the correct option is $B$.

(D) An electric dipole consists of two equal and opposite charges separated by a small distance $d$.
When placed in a non-uniform electric field,the electric field strength at the position of the two charges is different.
Since $F = qE$,the forces acting on the two charges are not equal in magnitude,resulting in a net linear force.
Additionally,because the forces act at different points,they create a torque about the center of the dipole.
Therefore,an electric dipole in a non-uniform electric field experiences both a net linear force and a torque.

(D) The electric field $E$ on the axis of a dipole at a distance $r$ is given by $E = \frac{2Kp}{r^3}$,where $K$ is Coulomb's constant and $p$ is the dipole moment.
The force $F$ on a charge $q$ is given by $F = qE = q \left( \frac{2Kp}{r^3} \right) = \frac{2Kpq}{r^3}$.
From this expression,we see that $F \propto \frac{1}{r^3}$.
If the distance is doubled,i.e.,$r' = 2r$,the new force $F'$ will be:
$F' = \frac{2Kpq}{(2r)^3} = \frac{2Kpq}{8r^3} = \frac{1}{8} \left( \frac{2Kpq}{r^3} \right) = \frac{F}{8}$.

(C) The dipole moment $p$ is defined as the product of the magnitude of one of the charges $q$ and the distance between them $2a$.
Given:
Charge $q = 3.2 \times 10^{-19} \ C$
Distance $2a = 2.4 \ \mathring{A} = 2.4 \times 10^{-10} \ m$
Electric field $E = 4 \times 10^5 \ V/m$ (Note: The electric field is extra information not required to calculate the dipole moment).
Formula: $p = q \times (2a)$
Calculation: $p = (3.2 \times 10^{-19} \ C) \times (2.4 \times 10^{-10} \ m)$
$p = 7.68 \times 10^{-29} \ C-m$.

(B) The dipole moment $p$ is given by $p = q \times (2a)$,where $q = 2 \times 10^{-6} \ C$ and $2a = 3 \times 10^{-2} \ m$.
$p = (2 \times 10^{-6}) \times (3 \times 10^{-2}) = 6 \times 10^{-8} \ Cm$.
The torque $\tau$ on a dipole in an electric field $E$ is given by $\tau = pE \sin \theta$.
For maximum torque,$\sin \theta = 1$,so $\tau_{max} = pE$.
$\tau_{max} = (6 \times 10^{-8} \ Cm) \times (2 \times 10^{5} \ N/C)$.
$\tau_{max} = 12 \times 10^{-3} \ Nm$.

(B) The electric field of a dipole at a point $(r, \phi)$ in polar coordinates is given by the radial component $E_r = \frac{2kp \cos \phi}{r^3}$ and the tangential component $E_t = \frac{kp \sin \phi}{r^3}$.
Here,$\phi = \pi/3$ is the angle the position vector $OP$ makes with the dipole axis (the $x$-axis).
The angle $\alpha$ that the resultant electric field makes with the radial vector $OP$ is given by $\tan \alpha = \frac{E_t}{E_r} = \frac{kp \sin \phi / r^3}{2kp \cos \phi / r^3} = \frac{1}{2} \tan \phi$.
Substituting $\phi = \pi/3$,we get $\tan \alpha = \frac{1}{2} \tan(\pi/3) = \frac{\sqrt{3}}{2}$,so $\alpha = \tan^{-1}\left(\frac{\sqrt{3}}{2}\right)$.
The angle $\theta$ that the electric field makes with the $x$-axis is the sum of the angle of the position vector $\phi$ and the angle $\alpha$,so $\theta = \phi + \alpha = \frac{\pi}{3} + \tan^{-1}\left(\frac{\sqrt{3}}{2}\right)$.

(A) The dipole moment $p$ is given by the product of the magnitude of one charge $q$ and the separation distance $2a$ (or $\ell$):
$p = q \times \ell$
Given:
$q = 10^{-10} \, \text{stC}$
$\ell = 1 \, \mathring{A} = 10^{-8} \, \text{cm}$
Substituting the values:
$p = 10^{-10} \, \text{stC} \times 10^{-8} \, \text{cm} = 10^{-18} \, \text{esu} \cdot \text{cm}$
Since $1 \, \text{Debye} = 10^{-18} \, \text{esu} \cdot \text{cm}$,
Therefore, $p = 1 \, \text{Debye}$.

(D) The electric field on the axis of a dipole at distance $x$ is given by $E_{\text{axis}} = \frac{1}{4\pi\epsilon_0} \cdot \frac{2p}{x^3}$.
The electric field on the equatorial line of a dipole at distance $y$ is given by $E_{\text{equatorial}} = \frac{1}{4\pi\epsilon_0} \cdot \frac{p}{y^3}$.
Given that $E_{\text{axis}} = E_{\text{equatorial}}$,we have:
$\frac{2p}{x^3} = \frac{p}{y^3}$.
Simplifying the equation:
$\frac{2}{x^3} = \frac{1}{y^3} \Rightarrow \frac{x^3}{y^3} = 2$.
Taking the cube root on both sides:
$\frac{x}{y} = \sqrt[3]{2} = 2^{1/3}$.
Therefore,the ratio $x:y$ is $\sqrt[3]{2}:1$.