Question

The magnitude of force due to an electric dipole of dipole moment $3.6\times {10}^{-29}\text{C-m}$ on an electron $25\text{nm}$ from the centre of the dipole along the dipole axis is

• A. $6.6\times {10}^{-15}\text{N}$
• B. $7.8\times {10}^{-15}\text{N}$
• C. $9.4\times {10}^{-15}\text{N}$
• D. $12\times {10}^{-15}\text{N}$

Answer 1

The correct option is A $6.6\times {10}^{-15}\text{N}$


Given:
$p=3.6\times {10}^{-29}\text{C-m};e=1.6\times {10}^{-19}\text{C}$

$x=25\times {10}^{-9}\text{m}$

We know that electric field due to electric dipole on axial line is given by

$E=\frac{2kpx}{(x^2-l^2{)}^2}$

in this case, $x>>l$

So, $E=\frac{2kp}{x^3}$

$\Rightarrow E=\frac{2\times 9\times {10}^9\times 3.6\times {10}^{-29}}{(25\times {10}^{-9}{)}^3}$

$\Rightarrow E=4.15\times {10}^4\text{N/C}$

Due to electric field $E$, force experience by the electron will be

$F=eE$

$\Rightarrow F=1.6\times {10}^{-19}\times 4.15\times {10}^4$

$\Rightarrow F=6.6\times {10}^{-15}\text{N}$

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