Question

Four charges $+q$ , $+q$ , $-q$ , and $-q$ are placed on $X-Y$ plane at the points whose coordinates are $\left(0.5,0\right)$ , $\left(0,0.5\right)$ , $\left(-0.5,0\right)$ and $\left(0,-0.5\right)$ respectively.

The electric field due to these charges at a point $P\left(r,r\right)$ , where $r>>0.5$ , will be

• A. $\frac{1}{4\pi {ϵ}_0}\times \frac{q}{2r^3}$
• B. $\frac{1}{4\pi {ϵ}_0}\times \frac{q}{r^3}$
• C. $\frac{1}{4\pi {ϵ}_0}\times \frac{3q}{r^3}$
• D. $\frac{1}{\pi {ϵ}_0}\times \frac{q}{r^3}$

Answer 1

Answer: (B)

The four charges can be considered two dipoles of equal magnitude and directed along the +x axis and +y axis.
The magnitude of dipole moment $=qx$
(distance between positive and negative charges on x-axis)
$p=q\times 1=q$
The resultant dipole moment of these two dipoles with be, $p_{\text{r}}=\sqrt{q^2+q^2}=q\sqrt{2}$
The direction of the resultant dipole moment will be at $45^{\circ }$ from the x-axis
The distance of the point $(r,r)$ from the dipole is $r\sqrt{2}m$
The resultant electric field is
$E=\frac{1}{4\pi {\epsilon }_0}\times \frac{2p_r}{(r\sqrt{2}{)}^3}$
$=\frac{1}{4\pi {ϵ}_0}\times \frac{2\times q\sqrt{2}}{2\sqrt{2}{r}^3}$
$=\frac{1}{4\pi {ϵ}_0}\times \frac{q}{r^3}$