Question

Two charges of $+25$ x ${10}^{-9}$ coulomb and $-25$ x ${10}^{-9}$ coulomb are placed $6$ m apart. Find the electric field intensity ratio at points $4$ m from the centre of the electric dipole
(i) on axial line
(ii) on equatorial line

• A. $\frac{1000}{49}$
• B. $\frac{49}{1000}$
• C. $\frac{500}{49}$
• D. $\frac{49}{500}$

Answer 1

The correct option is A $\frac{1000}{49}$

Case (i)
When point is on axial line
$E_{AC}=\frac{Kq}{r^2}$
$=\frac{K\times 2.5\times {10}^{-9}}{7^2}$
$E_{BC}=\frac{Kq}{r^2}$
$=\frac{K\times 2.5\times {10}^{-9}}{1^2}$
$\overrightarrow{E_1}=\overrightarrow{E_{BC}}-\overrightarrow{E_{AC}}$
$=\frac{K\times 2.5\times {10}^{-9}}{1^2}-\frac{K\times 2.5\times {10}^{-9}}{7^2}$
$=\frac{48\times K\times 2.5\times {10}^{-9}}{1^2}$............(1)
Case (ii)
When point is non equatorial line
$AC=\sqrt{A{B}^2+O{C}^2}$
$=\sqrt{3^2+4^2}$
$=5$ m
$\overrightarrow{E_2}=2\overrightarrow{E_{AC}}\cos \theta$
$=\frac{2\times K\times 25\times {10}^{-9}}{5^2}\times \frac{3}{5}$
$=\frac{6\times K\times 25\times {10}^{-9}}{125}$.........(2)
$\frac{E_2}{E_1}=\frac{48}{49}\times \frac{125}{6}=\frac{1000}{49}$