Question

Two charges $+ 20\, \mu C$ and $-20 \mu C$ are placed $10\, mm$ apart. The electric field at point $P$, on the axis of the dipole $10\, cm$ 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, $q -± 20 \mu C = ± 20 \times 10^{-6}\, C$, $2a = 10 \,mm = 10 \times 10^{-3}\, m$
$r = OP = 10 \,cm = 10 \times 10^{-2}\, m$
$\left|\vec{P}\right|=q\times2a=20\times10^{-6}\times10\times10^{-3}\,m=2\times10^{-7}\,m$
The electric field along $BP$, $\vec{E}=\frac{2\vec{p}r}{4\pi\varepsilon_{0}\left(r^{2}-a^{2}\right)^{2}}$
As $a < < r$,
$\vec{E}=\frac{2\left|\vec{p}\right|}{4\pi\varepsilon_{0}r^{3}}$
$=\frac{2\times2\times10^{-7}\times9\times10^{9}}{\left(10\times10^{-2}\right)^{3}}$
$=3.6\times10^{6}\,N\,C^{-1}$