Question

1. Consider a charge of -1 nC at (-1,0) m, and a charge of +1 nC at (1,0). Calculate the electric potential everywhere in the plane.

2. For the same geometry as in the first problem, calculate the electric field everywhere, using -?V =E. In 2 dimensions, ?V= (?subxV, ?subyV) = ?subxV x^ + ?suby Vy^

Answer 1

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.
  
  
  
  
  
• 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
  cos?=OC/OP = OC/a
  therefore OC=acos? similarly OD=acos?
  Now ,
  r₁ = QR?RD = OR-OD = r-acos?
  r₂ = PR?RC = OR+OC = r+acos?
  
  
  since magnitude of dipole is
  |p| = 2qa
  
  
  
• If we consider the case where r>>a then
  
  
  again since pcos?= p