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

The correct option is C $3.6\times {10}^{6}N{C}^{-1}$

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