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# Electric Potential Dipole

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

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)

 U(Dipole) = - pECosӨ = - p➝. E➝

## FAQs on Electric Potential Dipole

Question 1: What does (q1+ q2) = 0 Signify?

Answer: q1 + q2 = 0 q1 = - q2, is a system of two opposite point charges possessing equal magnitude, and this arrangement is called the electric dipole.

Question 2: What is an ideal Dipole?

Answer: For a given dipole moment viz: p = q x 2a, when the charge q gets larger and larger, the distance between the two charges keeps on decreasing. One must note that the product on the R.H.S should be managed to make it equal to the L.H.S. In this way we obtain an ideal dipole. However, the size of an ideal dipole is negligible.

Question 3: of 5 x 10-7C-m is aligned in a Uniform Eof 2 x 105N/m. Ascertain the dipole Potential Energy at 65° in the direction of the Electric Field.

Answer: Here, p= 5 x 10-7C-m, E= 2 x 105N/m, and Ө = 65°

Since,  U(Dipole) =  pE (CosӨi- CosӨf)

= 5 x 10-7 * 2 x 105* (Cos 0° - Cos 65°) = 10-1 * (1 - 0.422)

On solving, we get:

U(Dipole) = 5.8 x 10-2 J

Question 4: Can Electric Potential be Zero?