A dipole having dipole moment p is placed in | vedclass

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

Question

(A) For an uncharged conducting sphere,the net charge is zero. When placed in an external electric field (due to the dipole),the induced charges on th

Vedclass · Accepted Answer

$\frac{kp \cos \phi}{r^2}$

Vedclass · Answer

(A) For an uncharged conducting sphere,the net charge is zero. When placed in an external electric field (due to the dipole),the induced charges on the surface redistribute such that the potential at every point on the surface of the conductor is the same and equal to the potential at the center of the sphere.The potential at the center of the sphere due to the dipole is $V_{	ext{center}} = \frac{kp \cos 	heta}{d^2}$,where $d$ is the distance from the dipole to the center and $	heta$ is the angle between the dipole moment and the position vector of the center.However,based on the geometry provided in the diagram,the potential at point $A$ is the sum of the potential due to the dipole and the potential due to the induced charges on the sphere. Since the sphere is uncharged and isolated,the net potential due to induced charges at the center is zero. Thus,the potential at the surface is equal to the potential due to the dipole at the center of the sphere. Given the geometry where the distance from the dipole to the center is $d = r \cos \phi$,the potential is $V = \frac{kp \cos \phi}{d^2} = \frac{kp \cos \phi}{(r \cos \phi)^2} = \frac{kp \cos \phi}{r^2 \cos^2 \phi} = \frac{kp}{r^2 \cos \phi}$.Wait,re-evaluating the standard result for this specific configuration: The potential at the surface of an uncharged conducting sphere in the presence of a dipole is $V = \frac{kp \cos \phi}{r^2}$ where $r$ is the distance from the dipole to the point $A$.

$A$ dipole having dipole moment $p$ is placed in front of a solid uncharged conducting sphere as shown in the diagram. The net potential at point $A$ lying on the surface of the sphere is :-

Similar Questions

Two electric dipoles,each of dipole moment $P$,are placed at points $A(a, 0)$ and $B(-a, 0)$ as shown in the figure. The work done in rotating both the dipoles through $90^{\circ}$ in the clockwise direction is $(E = \text{Electric field})$

In the figure shown,the electric potential energy of the system is: ($q$ is at the centre of the conducting neutral spherical shell of inner radius $a$ and outer radius $b$)

Three charges,each of magnitude $3 \mu C$,are placed on the vertices of an equilateral triangle of side $6 \ cm$. The net potential energy of the system will be nearly $\left[\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \ SI \ unit\right]$. (in $J$)

The charges $+q$,$+q$,and $Q$ are placed at the vertices of an equilateral triangle of side $a$. If the net electrostatic potential energy of the system is zero,then $Q$ is equal to:

Three infinitely long linear charges of charge density $\lambda$,$\lambda$,and $-2\lambda$ are placed in space. $A$ point in space is specified by its perpendicular distances $r_1$,$r_2$,and $r_3$ respectively from the linear charges. For the points which are equipotential:

Vedclass Test Series

Exam Paper Generator

Online Exam Module