Question

Two charges each of $1\mu \text{C}$ but opposite in sign are $1\text{cm}$ apart. The electric field at a distance $10\text{cm}$ from the mid point on the axial line of this dipole is

• A. $18\times {10}^4\text{N/C}$
• B. $20\times {10}^4\text{N/C}$
• C. $36\times {10}^4\text{N/C}$
• D. $40\times {10}^4\text{N/C}$

Answer 1

The correct option is A $18\times {10}^4\text{N/C}$


Given:
$q=1\mu \text{C}={10}^{-6}\text{C};x=10\text{cm}={10}^{-1}\text{m}$

$2l=1\text{cm}={10}^{-2}\text{m}$

We know that dipole field intensity $E$ on the axial line of the dipole is given by

$E=\frac{2k|\overrightarrow{p}|x}{(x^2-l^2{)}^2}=\frac{2k\times q\times 2l\times x}{(x^2-l^2{)}^2}$

$\Rightarrow E=\frac{2\times 9\times {10}^9\times {10}^{-6}\times {10}^{-2}\times {10}^{-1}}{({10}^{-2}-0.25\times {10}^{-4}{)}^2}$

$\Rightarrow E=\frac{18\times {10}^4}{(1-0.0025{)}^2}$

$\Rightarrow E=18\times {10}^4\text{N/C}$

Hence, option $\left(a\right)$ is the correct answer.