Physics help

Question

An electric dipole consists of a pint charge -q at the point (0,a) and a point charge q at the pint(0,-a). Find the electric field at the point (x,0).

Answer 1

The electric field due to a charge q at a point distance r from it is given by:

 

E=kq/r²*ñ, where ñ is the unit vector pointing from the charge to that point.

 

In your case vector ñ₁, for the charge -q, is (x,-a)/√(x²+a²). For the charge q, vector ñ₂, is (x,a)/√(x²+a²). Distances from the charges q and -q are the same, r₁=r₂=√(x²+a²); Therefore, total electric field at a point (x,0) is given by the sum of the electric fields due to each charge, q and -q, respectively. (this is called superposition principle)

 

E=k(-q)/(x²+a²)*ñ₁+kq/(x2+a2)*ñ₂

 

E=kq/(x²+a²)*(ñ₂-ñ₁)

 

Plugging in expressions for ñ₂ and ñ₁, we obtain finally:

 

E=kq/(x²+a²)*(0,2a)/√(x²+a²);

 

Electric field points up in the positive y-direction. Its magnitude is:

 

E=2kqa/(x²+a²)^3/2