Question

Two charges $+20\mu C$ and $-20\mu C$ are placed $10mm$ apart. The electric field at point $P$, on the axis of the dipole $10cm$ away from its centre $O$ on the side of the positive charge is

• A. $8.6\times 10^{9}N{C}^{-1}$
• B. $4.1\times 10^{6}N{C}^{-1}$
• C. $3.6\times 10^{6}N{C}^{-1}$
• D. $4.6\times 10^{5}N{C}^{-1}$

Answer 1

Answer: (C)

Here, $q-\pm 20\mu C=\pm 20\times 10^{-6}C$, $2a=10mm=10\times 10^{-3}m$
$r=OP=10cm=10\times 10^{-2}m$
$\left|\vec{P}\right|=q\times 2a=20\times 10^{-6}\times 10\times 10^{-3}m=2\times 10^{-7}m$
The electric field along $BP$, $\vec{E}=\frac{2\vec{p}r}{4\pi {\epsilon }_0{\left(r^2-a^2\right)}^2}$
As $a<<r$,
$\vec{E}=\frac{2|\vec{p}|}{4\pi {\epsilon }_0{r}^3}$
$=\frac{2\times 2\times 10^{-7}\times 9\times 10^9}{{\left(10\times 10^{-2}\right)}^3}$
$=3.6\times 10^{6}N{C}^{-1}$