Question

Derive an expression for the electric field intensity at a point on the equatorial line of an electric dipole moment p and length 2a. What is the direction of this field?

Answer 1

Answer:

$\underline{\sf{\bold{Electrical\:field\:intensity\:due\: to\:electric\:dipole\:at\:a\:point\:on\:the\: equatorial\:line}}}$

$\setlength{\unitlength}{2.5cm}\begin{picture}(1.5, 1.5)\put(-1,0){\line(1,0){2}}\put(0,1){\line(-1,-1){1.01}}\put(-0.9,0.025){{y}}\put(0.8,0.025){{y}}\put(0,1){\line(1,-1){1.01}}\put(0,-0.1){{D}}\put(0,0){\line(0,1){1}}\put(-1.2,0){{+q}}\put(1,-0){{-q}}\put(1,-0.12){{B}}\put(-1,-0.12){{A}}\put(0,1){{P}}\put(-0.6,0.5){{x}}\put(0.6,0.5){{x}}\put(-0.1,-0.3){{2a}}\end{picture}$

$\sf{Consider\:a\:dipole\: and\:a\:point\:P\:on\: its\:equatorial\:line\:passing\:through\:centre\:of\:dipole.}\\\\\sf{Electrical\:field\:intensity\:at\:P\:due\: to\:+q\:Charge,}\\\\E_1=\dfrac{1}{4 \pi \epsilon_0}\times\dfrac{q}{x^2}\\\\\sf{Electrical\:field\:intensity\:at\:P\:due\:to\:-q\:Charge,}\\\\E_2=\dfrac{1}{4 \pi \epsilon_0}\times \dfrac{q}{x^2}\\\\Lets\:highlight\:E's\:direction\:at\:P.$$\setlength{\unitlength}{2cm}\begin{picture}(1,1)\put(0,0){\vector(1,0){1}}\put(0,0){\vector(0,1){1}}\put(0,0){\vector(0,-1){1}}\put(0,0){\vector(1,1){0.7}}\put(0,0){\vector(1,-1){0.7}}\put(0.7,0.7){{E1}}\put(1.1,0){{E1.cosy}}\put(1.1,-0.2){{E2.cosy}}\put(0.2,0.101){{y}}\put(0.2,-0.101){{y}}\put(-0.1,-1.2){{E2.siny}}\put(0.7,-0.8){{E2}}\put(-0.1,1.1){{E1.siny}}\end{picture}$

$\sf{Vertical\: Components =E_1siny=E_2siny\:\:(Cancel\: each\: other)}\\\\Horizontal\:components =E_1cosy=E_2cosy\\\\E_{total}=E_1cosy+E_2cosy\\\\E_{total}=2E_1cosy\\\\E_{total}=2\times \dfrac{1}{4 \pi \epsilon_0}\times \dfrac{q}{x^2}*cosy\\\\Again,\:We\:moves\: to\:figure.$

$\setlength{\unitlength}{2.5cm}\begin{picture}(1.5, 1.5)\put(-1,0){\line(1,0){2}}\put(0,1){\line(-1,-1){1.01}}\put(0,1){\line(1,-1){1.01}}\put(0,-0.1){{D}}\put(-0.091,0.4){{r}}\put(0,0){\line(0,1){1}}\put(-1.2,0){{+q}}\put(1,-0){{-q}}\put(1,-0.12){{B}}\put(-1,-0.12){{A}}\put(0,1){{P}}\put(-0.8,0.09){{y}}\put(-0.6,0.5){{x}}\put(-0.5,-0.15){{L}}\end{picture}\\\\\\x^2 =r^2+L^2\\\\\implies x^3 =\bigg(r^2+L^2\bigg)^{\dfrac{3}{2}}$

$\implies E_{total}=2\times \dfrac{1}{4 \pi \epsilon_0}\times \dfrac{q}{x^2}\times \dfrac{a}{x}\\\\E_{equt}=\dfrac{1}{4 \pi \epsilon_0}\times \dfrac{P}{x^3}\:\:...........( P=2aq)\\\\\\E_{equt}=\dfrac{1}{4 \pi \epsilon_0}\times \dfrac{P}{\bigg(r^2+ a^2\bigg)^{\dfrac{3}{2}}}.........................(a^2\:Can\:be\:neglected)\\\\\\\\\boxed{\bold{E_{equt}=\dfrac{1}{4 \pi \epsilon_0}\times \dfrac{P}{r^3}}}$