Question

If $W=\varphi +i\Psi$ represents the complex potential for an eelectric field.
Given $\Psi =x^2-y^2+\frac{x}{x^2+y^2}$, then the function $\varphi$ is

• A. $-2xy+\frac{y}{x^2+y^2}+C$
• B. $2xy+\frac{y}{x^2+y^2}+C$
• C. $-2xy+\frac{x}{x^2+y^2}+C$
• D. $2xy-\frac{y}{x^2+y^2}+C$

Answer 1

The correct option is A $-2xy+\frac{y}{x^2+y^2}+C$

$w=\phi +i\Psi$ Complex Potential and $\Psi =x^2-y^2+\frac{x}{x^2+y^2}$
Using Milne Thomson method.
Step 1: $\frac{\partial \Psi }{\partial x}=2x+\frac{y^2-x^2}{(x^2+y^2{)}^2}\approx {\psi }_2(x,y)$ (let)
Step 2: ${\psi }_2(z,0)=2z-\frac{1}{z^2}$
Step 3: $\frac{\partial \Psi }{\partial x}=-2y+\frac{-2xy}{(x^2+y^2{)}^2}\approx {\psi }_1(x,y)$ (let)
Step 4: ${\psi }_1(z,0)=0$
Step 5: $w=\int \left[{\varphi }_1\right(z,0)+{\varphi }_2(z,0\left)\right]dz+C$
$=\int [0+i(2z-\frac{1}{z^2})dz]+C$
$=i(z^2-\frac{1}{z})+C$
$=i(x^2-y^2=2ixy+\frac{1}{x+iy})+C$
$=(-2xy+\frac{y}{x^2+y^2}+C)+i(x^2-y^2+\frac{x}{x^2+y^2})$
$w=\varphi +i\Psi$
$\Psi =-2xy+\frac{y}{x^2+y^2}+C$