

We know that a dipole is an arrangement of two opposite and equal charges joined with a line passing through their centers and the distance between them is the dipole length.
Let’s suppose that an arbitrary point lies anywhere and at this point how much potential the following dipole creates is the thing which we need to find out.
[Fig.1: Image will be Uploaded Soon]
Since there are two charges, the electric potential becomes equal to the sum of potential produced by each of these point charges. In this article, we will learn to determine electric potential near a dipole.
Electric Potential in Dipole
We considered the point lying anywhere and determined its potential due to the two charges, so we are considering point P as the arbitrary point.
[Fig.2 : Image will be Uploaded Soon]
Now, to derive our required formula, we need to consider the point P as far away from the charges and the dipole length to be much smaller than the P’s distance.
Also, we can see that the potential across the different points will be different because the distance varies. As we all know that the potential and distance are related to each other.
Let’s say the distance between the center of the dipole and the point P is ‘r’. The dipole center is considered to overcome the priority issues for the positive and negative charge.
The dipole length is 2a, so the distance from charge +q to the center of the dipole is ‘a’, and the angle made by the line joining P to the center of the dipole is Θ.
From Fig.2, the potential due to charge +q is V1 and due to -q is V2, then the potential at point P will be the algebraic sum of these two potentials, which is:
VP = V1 + V2….(1)
We know that the potential due to a point charge (-q) is:
V = \[\frac{1}{4πε₀}\] \[\frac{-q}{r}\] ….(2)
Now, joining another line from point A on which -q charge lies to point P and name this distance as r2, we get equation (2) as:
V = \[\frac{1}{4πε₀}\] \[\frac{-q}{r₂}\] ….(3)
Now, due to point charge +q, and joining the line from point B on which +q lies and point B and name this distance as ‘r1’, we get the potential as:
V = \[\frac{1}{4πε₀}\] \[\frac{+q}{r₁}\] ] ….(4)
We know that the electric potential between dipole is the sum of the potential due to the two charges. Now, putting the values of (4) and (3) in (1), we get:
= \[\frac{q}{4πε₀}\] [\[\frac{-1}{r₁}\]+\[\frac{1}{r₂}\]] …..(5)
Now, to evaluate the for the unknown values viz: r1 and r2, we will be drawing a perpendicular in Fig.2, as shown below in Fig.3:
[Fig.3 : Image will be Uploaded Soon]
Fig.3: (Here, this point R is taken as point P according to our derivation, and R taken as P)
(One should note that the distances r, r1, and r2 are greater than the dipole length)
From Fig.3, PC = PO + OC
Also, ΔDQO and ΔRCO are congruent because ∠OCR = ∠QOD = 90° and ∠COR = ∠POQ = Θ
So, OC = OD
Now, to determine the value of OC from OCR, we have:
CosΘ = \[\frac{OC}{a}\]⇒ OC = aCosΘ….(6)
r2 ≈ PC = OP + OC = r + aCosΘ….(7)
Similarly,
r1 ≈ PD = PO - OD = PO - OC = r - aCosΘ…..(8)
Now, substituting the value of (7) & (8) in eq (5):
= \[\frac{q}{4πε₀}\] [\[\frac{-1}{(r+acosΘ)}\]+ \[\frac{1}{(r-acosΘ)}\]]
Solving this further:
= \[\frac{q}{4πε₀}\] [ \[\frac{(r+acosΘ)-(r-acosΘ)}{r^{2}-a^{2}cos^{2}Θ}\]]
= \[\frac{q}{4πε₀}\] [\[\frac{(2aCosΘ)}{(r^{2}-a^{2}Cos^{2}Θ)}\]]
= \[\frac{q*2a}{4πε₀}\] \[\frac{(2aCosΘ)}{(r^{2}-a^{2}Cos^{2}Θ)}\]
= \[\frac{p}{4πε₀}\] [\[\frac{(r+acosΘ)-(r-acosΘ)}{r^{2}-a^{2}cos^{2}Θ}\]]…(9)
(Since dipole moment, p = q x 2a)
(Here, we have not considered the polarity of a charge)
As we know that the distance ‘r’ is very large as compared to the dipole length ‘a’, so, ignoring the expression a2Cos2, we rewrite the equation (9) as:
= \[\frac{p}{4πε₀}\] \[\frac{cosΘ}{r^{2}}\]
This is the required equation for electric potential in a dipole.
Electric Potential Energy of A Dipole
[Image will be Uploaded Soon]
We know that in a dipole, equal and opposite forces cancel out each other, however, the torque acting parallel to the system rotates it.
If p➝ is the dipole moment, which is not aligned to the uniform electric field E➝
, then torque does the work in aligning E to p and the magnitude of this torque is:
τ = pESinΘ
Now, to rotate the dipole far from the alignment, we need to apply an external torque and perform some work. Also, this work can be stored in the form of potential energy or transform into other forms of energy. Now, the potential energy of an external field is:
pE(Dipole) = -pECosΘ
So, what happens here is, if the dipole is rotated from an initial angle Өi to Өf, then the potential energy is:
U(Dipole) = pE (CosӨi- CosӨf)...(10)
If the angle at which the potential energy is zero, then equation (10) becomes:
= pE (\[\frac{cosπ}{2}\] - CosӨ)
FAQs on Electric Potential Dipole
1. What is an electric potential dipole as described in the CBSE Class 12 Physics syllabus?
An electric potential dipole consists of two equal and opposite point charges separated by a small distance. It is a crucial concept in electromagnetism where the dipole moment (p) is defined as the product of charge and distance between the charges, directed from negative to positive. Understanding dipoles helps explain electric fields and potential in more complex arrangements.
2. How is the electric potential calculated at a point due to an electric dipole?
The electric potential V at a point due to a dipole is given by:
- For a point on the axial line: V = (1/4πε0) × (p cosθ/r2)
- For a point on the equatorial line: V = 0 (since contributions cancel out)
3. Why is the electric potential zero at the equatorial point of a dipole?
At the equatorial point of a dipole, the potentials due to both charges are equal in magnitude but opposite in sign. Thus, their effects cancel each other, resulting in a net electric potential of zero at this position.
4. What is meant by an ideal dipole, and how does it differ from a physical dipole?
An ideal dipole is a theoretical concept where the separation between charges approaches zero while the dipole moment (p = q × 2a) remains constant. In contrast, a physical dipole has a finite separation between charges. Ideal dipoles are used for simplifying calculations in Physics as per the CBSE 2025–26 syllabus.
5. How do you determine the potential energy of an electric dipole placed in a uniform electric field?
The potential energy (U) of an electric dipole in a uniform electric field (E) is given by U = -pE cosθ, where:
- p is the dipole moment
- θ is the angle between the dipole moment and the field direction
6. Why is understanding electric potential dipole important for CBSE board exams?
Electric potential dipole questions often test conceptual clarity and application skills. Board exams may ask about definitions, derivations, numerical problems, or real-life relevance. Mastering this topic strengthens understanding of fields and forces in Physics, which are frequently weighted in the CBSE Class 12 paper pattern.
7. What are the common misconceptions students have about electric dipole potential?
Common misconceptions include:
- Assuming the electric field and potential at the center of a dipole are both zero (only the field is zero, not the potential)
- Confusing the direction of dipole moment (it is from negative to positive charge)
- Believing the equatorial line has electric field zero (it is only the potential that is zero, not the field itself)
8. How do variations in dipole length or charge affect the electric potential produced?
Increasing the separation (2a) or the charge (q) directly increases the dipole moment (p = q × 2a), which increases the resulting electric potential and field intensity away from the dipole. Thus, both parameters are directly proportional to dipole strength, as per CBSE Physics concepts.
9. In what real-world situations do electric dipoles and their potential play a key role?
Electric dipole potential is vital in:
- Understanding behavior of water molecules (which have a natural dipole moment)
- Designing capacitors and sensors
- Studying molecular interactions in Chemistry and Biology

















