Question

A point dipole is located at the origin in some orientation. The electric field, at the point $(10cm,10cm)$ 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 $(20cm,20cm)?$

• 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 $(20,20)r^{\prime }$
$=\sqrt{20^2+20^2}=20\sqrt{2}$
$E^{\prime }=\frac{1}{4\pi t_0}\frac{\rho }{r^{{}^{\prime }3}}$
$E^{\prime }=\frac{1}{4\pi t_0}\times \frac{E\left(4\pi t_0{r}^3\right)}{r^{{}^{\prime }3}}$
$E^{{}^{\prime }}=\frac{E{r}^3}{r^{{}^{\prime }3}}=\frac{1\times 10^{-3}\times (0.1\sqrt{2}{)}^3}{(0.2\sqrt{2}{)}^3}$
$=1.25\times 10^{-4}\frac{v}{m}$