An electric dipole has charges of $+5.0\mu \text{C}$ and $-5.0\mu \text{C}$ separated by $5.0\text{cm}$. (a) Calculate the magnitude of the electric dipole moment. (b) Find the magnitude of the electric field at a point on the axial line of the dipole, $0.40\text{m}$ from the dipole centre. Use $\frac{1}{4\pi {\epsilon }_0}=9.0\times 10^9{\text{N⋅m}}^2{\text{/C}}^2$. (For a dipole with $p\ll r^2$, the axial field is $E=\frac{1}{4\pi {\epsilon }_0}\frac{2p}{r^3}$.)

Step-by-step solution

1. Write down the definition: dipole moment $p=q\times d$, where $q$ is the magnitude of either charge and $d$ is the separation.
2. Convert the separation to metres: $d=5.0\text{cm}=0.050\text{m}$. The charge magnitude is $q=5.0\mu \text{C}=5.0\times 10^{-6}\text{C}$.
3. Calculate the dipole moment: $p=(5.0\times 10^{-6})(0.050)=2.5\times 10^{-7}\text{C⋅m}$.
4. For part (b), identify the axial-line formula: $E=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{2p}{r^3}$, valid when $r\gg d$. Here $r=0.40\text{m}\gg 0.050\text{m}$, so the approximation is valid.
5. Compute $r^3=(0.40{)}^3=0.064{\text{m}}^3$. Then $2p=5.0\times 10^{-7}\text{C⋅m}$.
6. Substitute: $E=9.0\times 10^9\times \frac{5.0\times 10^{-7}}{0.064}=9.0\times 10^9\times 7.8125\times 10^{-6}$.
7. Final arithmetic: $E=70.3125\times 10^3\text{N/C}=7.03\times 10^4\text{N/C}$. Wait — let me recheck: $5.0\times 10^{-7}/0.064=7.8125\times 10^{-6}$. Multiply by $9.0\times 10^9$: $9.0\times 7.8125\times 10^3=70.3125\times 10^3=7.03\times 10^4\text{N/C}$. Hmm, let me re-examine. Actually $p=2.5\times 10^{-7}$, so $2p=5.0\times 10^{-7}$. $E=9.0\times 10^9\times \frac{5.0\times 10^{-7}}{0.064}$. The quotient $5.0\times 10^{-7}/0.064=7.8125\times 10^{-6}$. Then $9.0\times 10^9\times 7.8125\times 10^{-6}=70.3125\times 10^3=7.03\times 10^4\text{N/C}$. This seems large; let me double-check the dipole moment. $q=5\mu \text{C}$, $d=0.05\text{m}$, $p=2.5\times 10^{-7}\text{C⋅m}$. Yes. So $E\approx 7.0\times 10^4\text{N/C}$.
8. The field is directed along the dipole axis, from the $-q$ end toward the $+q$ end (i.e. away from the dipole's positive charge on the axial line).

Concept tested: Dipole moment and axial electric field

Published 22 September 2026 · How this solution was prepared and checked