(a) An electric dipole of dipole moment $\vec{p}$ consists of point charges $+q$ and $-q$ separated by a distance $2a$ apart. Deduce the expression for the electric field $\vec{E}$ due to the dipole at a distance $x$ from the centre of the dipole on its axial line in terms of the dipole moment $\vec{p}$. Hence show that in the limit $x \gg a$, $E \to \dfrac{2p}{4\pi\varepsilon_0 x^3}$. (b) Given the electric field in the region $\vec{E} = 2x\,\hat{i}$, find the net electric flux through the cube and the charge enclosed by it (the cube has side $a$ with one corner at the origin).

(a) $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2px}{(x^2 - a^2)^2} \to \dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{x^3}$ for $x \gg a$. (b) Net flux $\phi = 2a^3$; charge enclosed $q = 2\varepsilon_0 a^3$.

CBSE Class 12 · 2015 · Delhi · ElectrostaticsLong Answer

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(a) An electric dipole of dipole moment

p

consists of point charges

+ q

and

- q

separated by a distance

2 a

apart. Deduce the expression for the electric field

E

due to the dipole at a distance

x

from the centre of the dipole on its axial line in terms of the dipole moment

p

. Hence show that in the limit

x ≫ a

E \to \frac{2 p}{4 π ε _{0} x ^{3}}

.
(b) Given the electric field in the region

E = 2 x \hat{i}

, find the net electric flux through the cube and the charge enclosed by it (the cube has side

a

with one corner at the origin).

Solution

(a) Let P be an axial point at distance

x

; it is at

(x - a)

from

+ q

and

(x + a)

from

- q

E = \frac{1}{4 π ε _{0}} [\frac{q}{( x - a ) ^{2}} - \frac{q}{( x + a ) ^{2}}]

E = \frac{q}{4 π ε _{0}} \frac{( x + a ) ^{2} - ( x - a ) ^{2}}{( x - a ) ^{2} ( x + a ) ^{2}}

Using

(x + a)^{2} - (x - a)^{2} = 4 a x

and

(x - a) (x + a) = x^{2} - a^{2}

E = \frac{1}{4 π ε _{0}} \frac{2 p x}{( x ^{2} - a ^{2} ) ^{2}}, p = q (2 a)

For

x ≫ a

E \to \frac{1}{4 π ε _{0}} \frac{2 p}{x ^{3}}

(b)

E = 2 x \hat{i}

is along

x

, so only the faces normal to

x

carry flux.
Right face (

x = a

E = 2 a

, flux

= 2 a \cdot a^{2} = 2 a^{3}

. Left face (

x = 0

): flux

= 0

ϕ = 2 a^{3} \Rightarrow q_{enc} = ε_{0} ϕ = 2 ε_{0} a^{3}

CBSE Class 12 Board · 2015 · Delhi · Subjective

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CBSE Class 12 · 2015 · Delhi · ElectrostaticsLong Answer

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