The electric field in the xy-plane due to a point charge at (0,0) is a gradient

Question

The electric field in the xy-plane due to a point charge at (0,0) is a gradient field with a potential V(x,y) = k/((sqrt(x^2 + y^2)) where k > 0 is a physical constant.

a. find the oof the electric field in the x- and y- directions, where E(x,y) = - V (x,y).

b. Show the vectors of the electric field pint in the radial direction (utward form the region) and the radial component f E can be expressed as Er = k/r^2, where r = sqrt(x^2 + y^2)

c. othat the vector field is orthogonal to the equipotential curves at all the points in the domain of V.

Explanation / Answer

a)

V(x,y) = k/((sqrt(x^2 + y^2)

dV/dx=-3kx/(x^2 + y^2)^(3/2)

dV/dy=-3ky/(x^2 + y^2)^(3/2)

Electric field = - dV/dx i - dV/dy j

= 3kx/(x^2 + y^2)^(3/2) i +3ky/(x^2 + y^2)^(3/2) j

*****************************************************************************************

b)

Magnitude of electric field =(Ex^2+Ey^2)^0.5

=(3k/(x^2 + y^2)^(3/2))*(x^2 + y^2)^0.5

=3K/(x^2 + y^2)

******************************************************************************************

C)

Along equipotential curves,

k/((sqrt(x^2 + y^2))= constant

(x^2 + y^2)=k^2/c^2............which is a circle.

Since, the electric fields are radial lines,

We know that tangents to a circle are orthogonal to radial lines.

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