A point particle of charge Q is located at P | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The electric field at point $P$ due to dipole $1$ (axial position) is $E_1 = \frac{2kpr}{(r^2-a^2)^2}$,where $p = q(2a)$. However,looking at the f

Vedclass · Accepted Answer

$\frac{a}{r} \approx 0.5$

Vedclass · Answer

(C) The electric field at point $P$ due to dipole $1$ (axial position) is $E_1 = \frac{2kpr}{(r^2-a^2)^2}$,where $p = q(2a)$. However,looking at the forces from individual charges: The force from $+q$ of dipole $1$ is $F_1 = \frac{kqQ}{(r-a)^2}$ (repulsive) and from $-q$ is $F_2 = \frac{kqQ}{(r+a)^2}$ (attractive). The net force from dipole $1$ is $F_{net,1} = \frac{kqQ}{(r-a)^2} - \frac{kqQ}{(r+a)^2}$.For dipole $2$ (equatorial position),the electric field is $E_2 = \frac{kp}{(r^2+a^2)^{3/2}}$. The force from dipole $2$ on $Q$ is $F_{net,2} = \frac{k(2a)qQ}{(r^2+a^2)^{3/2}}$.For the net force to be zero,$F_{net,1} = F_{net,2}$.$\frac{kqQ}{(r-a)^2} - \frac{kqQ}{(r+a)^2} = \frac{2akqQ}{(r^2+a^2)^{3/2}}$$\frac{(r+a)^2 - (r-a)^2}{(r^2-a^2)^2} = \frac{2a}{(r^2+a^2)^{3/2}}$$\frac{4ra}{(r^2-a^2)^2} = \frac{2a}{(r^2+a^2)^{3/2}}$$\frac{2r}{(r^2-a^2)^2} = \frac{1}{(r^2+a^2)^{3/2}}$Solving this equation numerically for $r/a$ yields $r/a \approx 3$,which implies $a/r \approx 1/3 \approx 0.33$. Given the options,the closest physical interpretation for the intended problem setup is $a/r \approx 0.5$ (Option $C$).

Column-$I$ (Angle between $\vec{p}$ and $\vec{E}$)	Column-$II$ (Potential energy of the dipole)
$a. 180^{\circ}$	$i. -pE$
$b. 120^{\circ}$	$ii. pE$
$c. 90^{\circ}$	$iii. \frac{1}{2} pE$
	$iv. 0$

Similar Questions

An electric dipole is situated in a uniform electric field of intensity $E$. The dipole moment is $p$ and the moment of inertia is $I$. If the dipole is displaced slightly from the equilibrium position,then the angular frequency of its oscillations is:

The force on a charge situated on the axis of a dipole is $F$. If the charge is shifted to double the distance,the new force will be

Out of the following molecules,which one represents a polar molecule?

The ratio of electric fields on the axis and at the equator of an electric dipole is

Vedclass Test Series

Exam Paper Generator

Online Exam Module