Answer:
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)
