The electric field in the xy-plane due to an infinite line of charge along the z-axis is a gradient field with a potential function V(x, y)equals=c ln (ro/√ x²+y²)

where c > 0 is a constant and r0 is a reference distance at which the potential is assumed to be 0.

Required:
Find the components of the electric field in the x- and y-directions, where E(x, y) =∇V (x, y )

c(x,y)/(x²+y²)

Step-by-step explanation:

Since E(x,y) = -∇V (x,y ) and V(x,y) =c ln (ro/√ x²+y²)

Let ro/√(x²+y²) = u and √(x²+y²) = v

dV/du = c/u = c√(x²+y²)/ro,

du/dv =  -ro/(x²+y²) = -ro/² and dv/dx = x/√(x²+y²)

Using the chain rule,

So dV/dx = dV/du × du/dv × dv/dx

= c√(x²+y²)/ro × -ro/(x²+y²) × x/√(x²+y²)

dV/dx = -cx/(x²+y²)

- dV/dx = -(-cx/(x²+y²)) = cx/(x²+y²)

Also, dv/dy = y/√(x²+y²)

Using the chain rule

dV/dy = dV/du × du/dv × dv/dy

= c√(x²+y²)/ro × -ro/(x²+y²) × y/√(x²+y²)

dV/dy = -cy/(x²+y²)

- dV/dy = -(-cy/(x²+y²)) = cy/(x²+y²)

E(x,y) = -∇V (x,y )

= -(dV/dx)i + [-(dV/dy)]j

= [cx/(x²+y²)]i +[ cy/(x²+y²)]j

= c(x,y)/(x²+y²)

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step-by-step explanation: