The electric potential due to a small electric dipole at a large distance r from the center of the dipole is proportional
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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.
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 ad 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 p.
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.
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 ad 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 p.
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