Q: The electric potential at a point on the axis of an electric dipole depends on the distance \$r\$ of the point from the dipole as

The expression of electric potential for dipole is , \$V=\dfrac{1}{4\pi\epsilon_0} \dfrac{p\cos\theta}{r^2}\$ where \$p=qd\$ dipole moment. so,\$ V \propto \dfrac{1}{r^2} \$ Ans:(B)

Q: An electric field of \$1000\$V/m is applied to an electric dipole at angle of \$45^o\$. The value of electric dipole moment is \$10^{-29}\$C.m. What is the potential energy of the electric dipole?

\$U=-\vec{P}\cdot \vec{E}\$\$=-PE\cos\theta\$\$=-(10^{-29})(10^3)\cos 45^o\$\$=-0.707\times 10^{-26}\$J\$=-7\times 10^{-27}\$J.

Q: The electric potential at a point on the equatorial line of an electric dipole is

\$\textbf{Step 1: Electric potential at P due to dipole}\$\$\textbf{[Refer Fig.]}\$Let OP be equatorial line.Let \$r\$ be the distance of each charge from point P Potential due to charge \$q\$ at point \$P, V_1 = \dfrac {kq}{r}\$ Potential due to charge \$-q\$ at point \$P , V_2= \dfrac {-kq}{r}\$\$\textbf{Step 2: Total potential at point P}\$ \$V = V_{1} + V_{2}\$ \$\Rightarrow V = \dfrac {kq}{r} - \dfrac {kq}{r} = 0\$Hence, the net potential at a point on the equatorial line of an electric dipole is zero.Hence, Option (D) is correct.

Q: The electric potential due to a small electric dipole at a large distance \$r\$ from the centre of the dipole is proportional to :

The potential at a distance r due to dipole is \$\displaystyle V_{dip}(r)=\dfrac{1}{4\pi\epsilon_0} \dfrac{\vec{p}.\hat{r}}{r^2}\$ where \$\vec{p}=\$ dipole moment. Thus, \$V_{dip}(r) \propto \dfrac{1}{r^2}\$

Question 1

Draw the equipotential surfaces due to an electric dipole. Locate the points where the potential due to the dipole is zero.

Accepted Answer

Dipole consist of two charges of opposite nature and the same magnitude having a small distance between them. The axis between the two dipoles, perpendicular to the plane, is where the potential due to the dipole is zero (straight line joining the arrows between the two dipoles).

Question 2

The electric potential at a point on the axis of an electric dipole depends on the distance $r$ of the point from the dipole as

Accepted Answer

&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;The expression of&#10;electric potential for dipole is , $V=\dfrac{1}{4\pi\epsilon_0} \dfrac{p\cos\theta}{r^2}$&#10;where $p=qd$ dipole moment.&#10;&#10;so,$ V \propto \dfrac{1}{r^2} $&#10;&#10;Ans:(B)&#10;&#10;

Question 3

An electric field of $1000$V/m is applied to an electric dipole at angle of $45^o$. The value of electric dipole moment is $10^{-29}$C.m. What is the potential energy of the electric dipole?

Accepted Answer

$U=-\vec{P}\cdot \vec{E}$$=-PE\cos\theta$$=-(10^{-29})(10^3)\cos 45^o$$=-0.707\times 10^{-26}$J$=-7\times 10^{-27}$J.

Question 4

The value of electric potential at any point due to any electric dipole is:-

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Question 5

A sphere encloses an electric dipole within it. The total flux across the sphere is

Accepted Answer

The total electric flux coming out of the sphere is zero because the net charge enclosed by the surface is zero

Question 6

The electric potential at a point on the equatorial line of an electric dipole is

Accepted Answer

$\textbf{Step 1: Electric potential at P due to dipole}$$\textbf{[Refer Fig.]}$Let OP be equatorial line.Let $r$ be the distance of each charge from point P Potential due to charge $q$ at point $P, V_1 = \dfrac {kq}{r}$ Potential due to charge $-q$ at point $P , V_2= \dfrac {-kq}{r}$$\textbf{Step 2: Total potential at point P}$                 $V = V_{1} + V_{2}$                 $\Rightarrow V = \dfrac {kq}{r} - \dfrac {kq}{r} = 0$Hence, the net potential at a point on the equatorial line of an electric dipole is zero.Hence, Option (D) is correct.

Question 7

Two point dipoles of dipole moment $\vec {p}_{1}$ and $\vec {p}_{2}$ are at a distance $x$ from each other and $\vec {p}_{1}\parallel \vec {p}_{2}$. The force between the dipoles is

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Question 8

Potential due to an electric dipole is represented by:

(where $p$ is the dipole moment, $r (> > 2 a)$ is the distance at which the potential $V$ of the dipole is calculated and $θ$ is the angle between the distance vector and the dipole.)

Accepted Answer

We resolve the electric dipole into horizontal and vertical components as shown in the figure.Parallel component of electric field $E_{\parallel} = \dfrac{2p \cos\theta}{4\pi \epsilon_o r^3}$Perpendicular of electric field $E_{\perp} = \dfrac{p\sin \theta}{4\pi \epsilon_o r^3}$Potential due to dipole $V = -\int \vec{E_{\parallel}} .\vec{dr} - \int \vec{E_{\perp}}. \vec{dr}$Second term goes to zero as $\vec{E_{\perp}}$ is perpendicular to $\vec{dr}$.Thus potential due to dipole $V = -\dfrac{2p \cos\theta}{4\pi \epsilon_o } \int \dfrac{dr}{r^3}$Or  $V = -\dfrac{2p \cos\theta}{4\pi \epsilon_o } \times \dfrac{1}{(-2) r^2}$$\implies$ $V = \dfrac{p \cos\theta}{4\pi \epsilon_o r^2} $

Question 9

The electric potential due to a small electric dipole at a large distance $r$ from the centre of the dipole is proportional to :

Accepted Answer

&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;The potential at a&#10;distance r due to dipole is $\displaystyle V_{dip}(r)=\dfrac{1}{4\pi\epsilon_0}&#10; \dfrac{\vec{p}.\hat{r}}{r^2}$ where $\vec{p}=$ dipole moment.&#10;&#10;Thus, $V_{dip}(r)&#10;\propto \dfrac{1}{r^2}$&#10;&#10;

Question 10

The equipotential surface of an electric dipole is :

Accepted Answer

Equipotential surface is a surface over which the potential due to any charge configuration is same throughout. The equipotential surfaces of an electric dipole are shown in the figure. Since the distance of each point on a plane passing through the centre of the electric dipole and perpendicular to the axis of the electric dipole is equal, the potential throughout the plane is zero. Hence it is one possible equipotential surface.

Potential due to Elelctric Dipole

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Potential Due to an Electric Dipole

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