Question

Obtain an expression for electric potential due to an electric dipole at a point whose position vector makes an angle theta with dipole

Answer 1

EXPLANATION

Potential due to an electric dipole

We already know that electric dipole is an arrangement which consists of two equal and opposite charges +q and -q separated by a small distance 2a.

Electric dipole moment is represented by a vector p of magnitude 2qa and this vector points in direction from -q to +q.

To find electric potential due to a dipole consider charge -q is placed at point P and charge +q is placed at point Q as shown below in the figure.

Potential due to an electric dipole

Since electric potential obeys superposition principle so potential due to electric dipole as a whole would be sum of potential due to both the charges +q and -q. Thus

where r1 and r2 respectively are distance of charge +q and -q from point R.

Now draw line PC perpandicular to RO and line QD perpandicular to RO as shown in figure. From triangle POC

cosθ=OC/OP = OC/a

therefore OC=acosθ similarly OD=acosθ

Now ,

r1 = QR≅RD = OR-OD = r-acosθ

r2 = PR≅RC = OR+OC = r+acosθ

since magnitude of dipole is

|p| = 2qa

If we consider the case where r>>a then

again since pcosθ= p·rˆ where, rˆ is the unit vector along the vector OR then electric potential of dipole is

for r>>a

From above equation we can see that potential due to electric dipole is inversly proportional to r2 not 1/r which is the case for potential due to single charge.

Potential due to electric dipole does not only depends on r but also depends on angle between position vector r and dipole moment

Answer 2

Answer:

The amount of labor required to shift a unit charge from a reference point to a specific place in an electric field is known as Electric Potential.

Explanation:

Allow an electric dipole to be made up of two equal and opposite point charges, –q at A and +q at B, separated by a little distance AB = 2a, with the center at O.

p = q × 2a is the dipole moment

At any point P, where OP=r and ∠BOP=θ, we will calculate potential

Let BP=r₁ and AP=r₂

Let's draw AC perpendicular PQ and BD perpendicular PO

In ΔAOC cosθ = OC/OA = OC/a

OC = acosθ

Similarly, OD = acosθ

Potential at P due to +q = $\frac{1}{4\pi e0}$ $\frac{q}{r2}$

Potential at P due to −q = $\frac{1}{4\pi e0}$ $\frac{q}{r1}$

Net potential at P due to the dipole V = $\frac{1}{4\pi e0}$($\frac{q}{r2}$ - $\frac{q}{r1}$)

⇒ V = $\frac{q}{4\pi e0}$($\frac{1}{r2}$ - $\frac{1}{r1}$)

Now, r₁ = AP = CP

= OP + OC

=r + acosθ

And r₂ = BP = DP

= OP – OD

=r − acosθ

V = $\frac{q}{y4\pi e0}$($\frac{1}{r - acos0}$ - $\frac{1}{r + acos0}$)

= $\frac{q}{4\pi e0}$($\frac{2acos0}{r^{2} - a^{2} cos^{2} 0 }$)

= $\frac{pcos0}{r^{2} - a^{2}cos^{2} 0 }$(since p = 2aq)

Thus due to an electric dipole, potential, V ∝ $\frac{1}{r^{2} }$. The electric potential of an electric dipole is 0 at all points along its equatorial line.