Electric Dipole
If two charges of equal magnitude and opposite nature are present at a very small distance from each other, they form an electric dipole.
If two charges +q and - q are situated at a distance `2 l` (may be 2r or 2a).
`star` The midpoint between the two charges is called the center of electric dipole.
`star` The line joining these charges is called axial line.
`star` The line perpendicular to axial line and passing through the center is called equatorial line.
Electric Dipole Moment
The product of the charge and the distance between them is called the dipole moment.
If we consider the distance 2a between the charges to be a vector whose direction is from -q to +q.
Then, from definition,
`\vec p = q (2\vec a)`
Unit of electric dipole moment = Coulomb \times meter
Dimensions of electric dipole moment = ` [M^0L^1T^1A^1]`
Examples of Electric Dipole
HCl, Nacl etc.
Q. The distance between `Na^+` and `Cl^{-1}` ions in Nacl is 1.28 Ã…. Calculate the electric dipole moment of the molecule. (Ans. 2048 ` \times 10^{-29}`C m)
Electric Field Due to Electric Dipole
We will find the electric field on the axial line and equatorial line of the electric dipole.
Electric Field due to Dipole at Axial Point
Question: Derive an expression for an electric field due to an electric dipole at a point on its axial line.
Consider an electric dipole placed on the x-axis with charges +q on A and - q on B separated by 2a distance. The Midpoint of the electric dipole is O. A point C is located at a distance of r from the midpoint O of the dipole on the axial line where we have to calculate electric field.
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| Electric Field due to Dipole at Axial Point |
Electric field at point C due to charge - q
`E_A = \frac{K q}{(AC)^2}`
`E_A = \frac{K q}{(r + a)^2}` `{\text{direction}\vec{CA}}`
Electric field at point C due to charge + q
`E_B = \frac{K q}{(BC)^2}`
`E_B = \frac{K q}{(r - a)^2}` `{\text{direction}\vec{BC}}`
Resultant electric field at point C.
`E_C = E_B - E_A`
`E_C =`` \frac{K q}{(r - a)^2}` - `\frac{K q}{(r + a)^2}`
`E_C =`` Kq[\frac{1}{(r - a)^2}` - `\frac{1}{(r + a)^2}]`
`E_C = Kq[ \frac{(r + a)^2 - (r - a)^2}{(r - a)^2(r + a)^2}]`
`E_C = Kq[ \frac{(r^2 + a^2 +2ar) - (r^2 + a^2 - 2ar)}{{(r - a)(r + a)}^2}]`
`E_C = Kq[\frac{r^2 + a^2 +2ar - r^2 - a^2 + 2ar}{(r^2 - a^2)^2}]`
`E_C = Kq[\frac{4ar}{(r^2 - a^2)^2}]`
`E_C = \frac{2 K q (2a) r}{(r^2 - a^2)^2}`
`E_C = \frac{2 K p r}{(r^2 - a^2)^2}` `{\because q\times2a = p}`
Vector form of electric field on axial line due to electric dipole
`\vec E_C = \frac{2 K \vec {p} r}{(r^2 - a^2)^2}`
If the value of a is very small as compared to r (r>>>a). Then `a^2` can be considered negligible in comparison with `r^2` Then, `(r^2 - a^2)^2` can be written as `r^4`.
Thus,
`E_C = \frac{2 K p r}{(r^2)^2}`
`E_C = \frac{2 K p r}{r^4}`
`E_C = \frac{2 K p }{r^3}`
Vector form of electric field on axial line due to electric dipole
`\vec E_C = \frac{2 K \vec p }{r^3}`
Conclusion
`\star` The intensity does not depends upon `r^{-1}` as a single charge does, whereas it depends upon `r^{-3}`.
`\star` The electric field intensity decreases rapidly with distance.
`\star` On axial line, the direction of the electric field is along the electric dipole moment.
Frequently Asked Questions (FAQs) - Electric Dipole - Definition, Formula, FAQs
Question: What is the dipole moment's SI unit?
Answer: SI unit for dipole moment is Coulomb `\times m`.
Question: What is the dipole moment's dimension?
Answer: Dimension for dipole moment is ` [M^0L^1T^1A^1]`.
Question: What is an electric dipole?
Answer: If two charges of equal magnitude and opposite nature are present at a very small distance from each other, they form an electric dipole.
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