Problem — Electric dipole

Problem

Two charges $+q$ and $-q$ (with $q = 2\ \mu\text{C}$) are separated by a distance $d = 3\ \text{mm}$. Calculate the magnitude of the dipole moment $p$ and the electric field on the dipole axis, at $20\ \text{cm}$ from the centre.

Solution

Dipole moment. By definition $p = q\,d$:

$p = q\,d = (2\cdot 10^{-6})(3\cdot 10^{-3}) = 6\cdot 10^{-9}\ \text{C}\cdot\text{m}$

Field on the axis (approximation $r \gg d$). Since $r = 0{,}20\ \text{m} \gg d = 3\ \text{mm}$, the axial dipole field formula holds:

$E_{\text{ass}} = \frac{2kp}{r^3} = \frac{2\cdot 8{,}99\cdot 10^9 \cdot 6\cdot 10^{-9}}{(0{,}20)^3} = \frac{107{,}88}{8\cdot 10^{-3}} = 1{,}35\cdot 10^4\ \text{V/m}$

The field on the axis points in the same direction as the dipole moment $\vv p$, i.e. from the $-q$ charge towards the $+q$ charge.

$\ev{p = 6\cdot 10^{-9}\ \text{C}\cdot\text{m},\ \ E_{\text{ass}}\approx 1{,}35\cdot 10^4\ \text{V/m}}$

Links

Topics: Electric field and potential Concepts: Electric dipole · Electric field