Question

Two charges of $+25\times {10}^{-9}$ coulombs and $-25\times {10}^{-9}$ coulomb are placed 6m apart. Find the electric field intensity ratio at points 4m from the centre of the electric dipole.
(i) On axial line
(ii) On equatorial line
(A) ${\displaystyle \frac{1000}{49}}$
(B) ${\displaystyle \frac{49}{1000}}$
(C) ${\displaystyle \frac{500}{49}}$
(D) ${\displaystyle \frac{49}{500}}$

Answer 1

Hint: To find the electric field intensity an axial line or equatorial line due to +q and –q charge we use the point charge formula i.e., $E={\displaystyle \frac{kq}{r^2}}$ and after then acceleration to the direction of electric field we can find the complete solution for axial & equatorial.

Complete step by step solution:
We have to calculate the electric field intensity on the axial line and on the equatorial line.
Case-I
For axial line

The electric field intensity due to point charge $-q={\displaystyle \frac{-kq}{r^2}}$
For –q charge r is the distance between A & C i.e., $=4+3=7m$
$E_q={\displaystyle \frac{-9\times {10}^9\times 25\times {10}^{-9}}{7\times 7}}={\displaystyle \frac{-25\times 9}{49}}$
Now, the electric field intensity due to point charge +q is $={\displaystyle \frac{kq}{r^2}}$
Here r is distance between B & C i.e., $4-3=1m$
$E_{+q}={\displaystyle \frac{9\times {10}^9\times 25\times {10}^{-9}}{1\times 1}}=25\times 9$
Hence the electric field intensity at point C i.e., on axial line is $E_a=E_{+q}+E_{-q}$
$=25\times 9-{\displaystyle \frac{25\times 9}{49}}=25\times 9\left(1-{\displaystyle \frac{1}{49}}\right)$
$E_a=\left({\displaystyle \frac{48}{49}}\right)(25\times 9)$
$E_a={\displaystyle \frac{10,800}{49}}$ ……(i)

Case-II
If point C is situated on equatorial line then electric field intensity can be calculated as

Here magnitude of electric field intensity for +q & -q are same i.e., $E={\displaystyle \frac{kq}{r^2}}$
Here r is distance between +q and –q to point C i.e., AC & BC which are equal
So, $AC=BC=r=\sqrt{{(3)}^2+{(4)}^2}$
$r=5m$
Then, $E_{+q}=E_{-q}=E={\displaystyle \frac{9\times {10}^9\times 25\times {10}^{-9}}{5\times 5}}$
$E={\displaystyle \frac{25\times 9}{25}}=9{\displaystyle \frac{N}{C}}$
Now, the total electric field intensity at point C i.e., on equatorial line according to diagram is given as –
$E_{eq}=E_{+q}\cos \theta +E_{-q}\cos \theta$
$=2E\cos \theta$
From diagram $\cos \theta ={\displaystyle \frac{3}{5}}$
$E_{eq}=2\times 9\times {\displaystyle \frac{3}{5}}={\displaystyle \frac{54}{5}}$ …..(2)
From equation (1) & (2) we can easily calculate the ratio of $E_a$ & $E_{eq}$
So, ${\displaystyle \frac{(1)}{(2)}}$
${\displaystyle \frac{E_a}{E_{eq}}}={\displaystyle \frac{10,800}{49}}\times {\displaystyle \frac{5}{54}}$
${\displaystyle \frac{54,000}{49\times 54}}$
So, ratio of $E_a$ & $E_{eq}$
$E_a:E_{eq}=1000:49$

So, the correct answer is (A) ${\displaystyle \frac{1000}{49}}$

Note: In this type of question basically we use electric field intensity due to point charge separately and then add them. But in many cases of dipole numericals $a<<<r$ i.e., distance from the centre of dipole is very large comparatively the distance between the 2 charges. So, in this case we can directly use 2 formulas which are given as
$E_a={\displaystyle \frac{2kp}{r^3}}$
$E_{eq}={\displaystyle \frac{kr}{r^3}}$
Where p is dipole moment of electric dipole i.e., $p=qa$