Question 2
(a) Given that A ~ = (x + 2y + 4z)ˆ
i + (2x - 3y- z)ˆj +(4x-y + 2z)kˆ
(i). Show that the vector field A ~ is irrotational
(ii). Find the scalar potential φ such that A~ = ∇φ, if φ(0, 0, 0) = 1
1)
∇×A⃗=0⃗,\nabla ×\vec A=\vec 0,∇×A=0,
∇×A⃗=(∂Az∂y−∂Ay∂z)i⃗+(∂Ax∂z−∂Az∂x)j⃗+(∂Ay∂x−∂Ax∂y)k⃗=(−1+1)i⃗+(4−4)j⃗+(2−2)k⃗=0⃗,\nabla×\vec A=(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z})\vec i+(\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x})\vec j+(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y})\vec k=(-1+1)\vec i+(4-4)\vec j+(2-2)\vec k=\vec 0,∇×A=(∂y∂Az−∂z∂Ay)i+(∂z∂Ax−∂x∂Az)j+(∂x∂Ay−∂y∂Ax)k=(−1+1)i+(4−4)j+(2−2)k=0,
2)
∇φ=∂A∂xi⃗+∂A∂yj⃗+∂A∂zk⃗=i⃗−3j⃗+2k⃗.\nabla \varphi=\frac{\partial A}{\partial x}\vec i+\frac{\partial A}{\partial y}\vec j+\frac{\partial A}{\partial z}\vec k=\vec i-3\vec j+2\vec k.∇φ=∂x∂Ai+∂y∂Aj+∂z∂Ak=i−3j+2k.
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