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 at the dipole
• D. $1.25\times {10}^{-4}$ V/m

Answer 1

The correct option is D $1.25\times {10}^{-4}$ V/m

at point (10,10) $E={10}^{-3}\frac{v}{m}$

so its distance from the origin $R=\sqrt{{10}^2+{10}^2}=10\sqrt{2}cm$

$E=\frac{1}{4\pi t_0}\frac{\rho }{r^3}⟹\rho =4\pi t_0{r}^3\left(E\right)$

the electric field at (20,20)$r^{\prime }=\sqrt{{20}^2+{20}^2}=20\sqrt{2}$

$E^{\prime }=\frac{1}{4\pi t_0}\frac{\rho }{r^{\prime 3}}$

$E^{\prime }=\frac{1}{4\pi t_0}\times \frac{E\left(4\pi t_0{r}^3\right)}{r^{\prime 3}}$

$E^{\prime }=\frac{E{r}^3}{r^{\prime 3}}=\frac{1\times {10}^{-3}\times {\left(0.1\sqrt{2}\right)}^3}{{\left(0.2\sqrt{2}\right)}^3}=1.25\times {10}^{-4}\frac{v}{m}$