Question

A point dipole is located at the origin in some orientation. The electric field, at the point $(10\, cm , 10 \, cm )$ on the $x-y$ plane is measured to have a magnitude $1.0 \times 10^{-3}\, V / m$. What will be the magnitude of the electric field at the point $(20\, cm , 20 \, cm ) ?$

• A. $ 5.0\times 10^{-4}\,V/m $
• B. $ 2.5\times 10^{-4}\,V/m $
• C. It will depend on the orientation of the dipole
• D. $ 1.25\times 10^{-4}\,V/m $

Answer 1

Answer: (D)

at point $(10,10) E =10^{-3} \frac{ v }{ m }$
so its distance from the origin $R =\sqrt{10^{2}+10^{2}}=10 \sqrt{2} \,cm$
$E =\frac{1}{4 \pi t _{0}} \frac{\rho}{ r ^{3}} $
$\Rightarrow \rho=4 \pi t _{0} r ^{3}( E )$
the electric field at $(20,20) r '$
$=\sqrt{20^{2}+20^{2}}=20 \sqrt{2}$
$E '=\frac{1}{4 \pi t _{0}} \frac{\rho}{ r ^{' 3}}$
$E'=\frac{1}{4 \pi t _{0}} \times \frac{ E \left(4 \pi t _{0} r ^{3}\right)}{ r ^{'3}}$
$E ^{'}=\frac{ Er ^{3}}{ r ^{'3}}=\frac{1 \times 10^{-3} \times(0.1 \sqrt{2})^{3}}{(0.2 \sqrt{2})^{3}}$
$=1.25 \times 10^{-4} \frac{ v }{ m }$