Question 1

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

The expression of
electric potential for dipole is ,  V=\dfrac{1}{4\pi\epsilon_0} \dfrac{p\cos	heta}{r^2} 
where  p=qd  dipole moment.

so,  V \propto \dfrac{1}{r^2}

Ans:(B)

Question 2

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 3

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

Accepted Answer

extbf{Step 1: Electric potential at P due to dipole} 	extbf{[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} 	extbf{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 4

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 5

If a point lies at a distance  'x'  from the mid point of the dipole along its axis, the electric potential at this point is proportional to :

Accepted Answer

The electric potential of dipole is given by:   V= \dfrac{p \space cos 	heta}{4 \pi \epsilon_0 \space r^2} where  r  is position vector with respect to mid point of dipole and  	heta  is angle between  r  and  p  (dipole moment). Therefore, electric potential at a point which lies at a distance ′x′′x′ from the mid point of the dipole will be proportional to  \dfrac{1}{x^2} .

Question 6

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	heta =-(10^{-29})(10^3)\cos 45^o =-0.707	imes 10^{-26} J =-7	imes 10^{-27} J.

Question 7

Figure shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on r for r/a >> 1, and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).

Accepted Answer

Four charges of same magnitude are placed at points X, Y, Y, and Z respectively, as shown in the following figure.A point is located at P, which is r distance away from point Y.The system of charges forms an electric quadrupole.It can be considered that the system of the electric quadrupole has three charges.Charge +q placed at point XCharge −2q placed at point YCharge +q placed at point Z XY=YZ=a YP=r PX=r+a PZ=r-a Electrostatic potential caused by the system of three charges at point P is given by, \displaystyle V=\frac{1}{4\pi\in_0}\left[\frac{q}{XP}-\frac{2q}{YP}+\frac{q}{ZP} ight] \displaystyle =\frac{1}{4\pi\in_0}\left[\frac{1}{r+a}-\frac{2q}{r}+\frac{q}{r-a} ight] \displaystyle =\frac{q}{4\pi\in_0}\left[\frac{r(r-a)-2(r+a)(r-a)+r(r+a)}{r(r+a)(r-a)} ight] =\displaystyle \frac{q}{4\pi\in_0}\left[\frac{r^2-ra-2r^2+2a^2+r^2+ra}{r(r^2-a^2)} ight] \displaystyle =\frac{q}{4\pi\in_0}\left[\frac{2a^2}{r(r^2-a^2)} ight] =\displaystyle \frac{2qa^2}{4\pi\in_0r^3\left(1-\frac{a^2}{r^2} ight)} Since \displaystyle \frac{r}{a} > > 1 , herefore \displaystyle \frac{a}{r} < < 1 \displaystyle \frac{a^2}{r^2} is taken as negligible. herefore V\displaystyle =\frac{2qa^2}{4\pi\in_0r^3} It can be inferred that potential, \displaystyle V \propto\frac{1}{r^3} However, it is known that for a dipole, \displaystyle V \propto\frac{1}{r^2} And, for a monopole, \displaystyle V \propto\frac{1}{r}

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.)

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: 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 due to an electric dipole at a distance r from the centre of the dipole and at an angle \theta to the dipole axis is ( p is the dipole moment, 2a is the length of the dipole) :

From figure, cos\theta = \dfrac{r^2 + a^2 - r_+^2}{2ra} We get r_+ = (r^2 + a^2 - 2ra \cos\theta)^{1/2} As r>>a , we can neglect a^2 as compared to r^2 \therefore r_+ = (r^2 - 2ra \cos\theta)^{1/2} Or r_+ = r (1 - \dfrac{2a}{r} \cos\theta)^{1/2} Using (1+x)^n = 1+nx for x<<1 We get r_+ = r (1 - \dfrac{a}{r} \cos\theta) \implies r_+ = r - a \cos\theta Similarly, r_- = r + a \cos\theta Electric potential at point A V = V_q + V_{-q} \therefore V = \dfrac{q}{4\pi \epsilon_o r_+} - \dfrac{q}{4\pi \epsilon_o r_-} Or V = \dfrac{q}{4\pi \epsilon_o }[\dfrac{1}{r_+} - \dfrac{1}{ r_-}] Or V = \dfrac{q}{4\pi \epsilon_o }[\dfrac{1}{r-a\cos\theta} - \dfrac{1}{ r+ a\cos\theta}] Or V = \dfrac{q}{4\pi \epsilon_o }[\dfrac{2a \cos\theta}{(r-a\cos\theta)(r+a\cos\theta)}] \implies V = \dfrac{p \cos\theta}{4\pi\epsilon_o(r^2-a^2\cos^2\theta)}

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	heta}{4\pi \epsilon_o r^3} Perpendicular of electric field  E_{\perp} = \dfrac{p\sin 	heta}{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	heta}{4\pi \epsilon_o } \int \dfrac{dr}{r^3} Or   V = -\dfrac{2p \cos	heta}{4\pi \epsilon_o } 	imes \dfrac{1}{(-2) r^2} \implies   V = \dfrac{p \cos	heta}{4\pi \epsilon_o r^2}

Question 9

The electric potential due to an electric dipole at a distance $r$ from the centre of the dipole and at an angle $θ$ to the dipole axis is ( $p$ is the dipole moment, $2 a$ is the length of the dipole) :

Accepted Answer

From figure, cos heta = \dfrac{r^2 + a^2 - r_+^2}{2ra} We get r_+ = (r^2 + a^2 - 2ra \cos heta)^{1/2} As r>>a , we can neglect a^2 as compared to r^2 herefore r_+ = (r^2 - 2ra \cos heta)^{1/2} Or r_+ = r (1 - \dfrac{2a}{r} \cos heta)^{1/2} Using (1+x)^n = 1+nx for x<<1 We get r_+ = r (1 - \dfrac{a}{r} \cos heta) \implies r_+ = r - a \cos heta Similarly, r_- = r + a \cos heta Electric potential at point A V = V_q + V_{-q} herefore V = \dfrac{q}{4\pi \epsilon_o r_+} - \dfrac{q}{4\pi \epsilon_o r_-} Or V = \dfrac{q}{4\pi \epsilon_o }[\dfrac{1}{r_+} - \dfrac{1}{ r_-}] Or V = \dfrac{q}{4\pi \epsilon_o }[\dfrac{1}{r-a\cos heta} - \dfrac{1}{ r+ a\cos heta}] Or V = \dfrac{q}{4\pi \epsilon_o }[\dfrac{2a \cos heta}{(r-a\cos heta)(r+a\cos heta)}] \implies V = \dfrac{p \cos heta}{4\pi\epsilon_o(r^2-a^2\cos^2 heta)}

Question 10

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

Accepted Answer

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