The image contains a graph with a rectangular shape. The rectangle is aligned with the x and y axes.- The x-axis is labeled from 0 to \( b \), and the y-axis is labeled from 0 to \( a \).- The bottom side of the rectangle along the x-axis has \( V = 0 \).- The left side of the rectangle along the y-axis at \( x = 0 \) has \( V = V_0 \) at \( y_0 \).- The top side of the rectangle has \( V = 0 \).- There is a "Gap" noted on the top right side of the rectangle.- The right side of the rectangle has \( V = 0 \).Below the diagram, there is a problem statement:a) Use the method of separation of variables to find \( V(x, y) \) inside the rectangle.

The potential function V(x,y) satisfies the Laplace equation,Vxx+Vyy=0, 0≤x≤b , 0≤y≤aFrom the given picture, the conditions areV(x,0)=0=V(x,a)V(0,y)=V0, V(b,y)=0Using separation of variables method:Let V(x,y)=X(x)Y(y)From the conditions on y, V(x,0)=X(x)Y(0)=0V(x,a)=X(x)Y(a)=0Thus Y(0)=Y(a)=0From the equation ,X″Y+XY″=0X″X=−Y″Y=λ (seperation constant)Thus two ODEs are x″=λX , Y″=−λYThe eigenvalue problem:Y″+λY=0, Y(0)=Y(a)=0Case-1: λ<0. λ=−p2, p≠0Y″=p2Y⇒ Y(y)=Acosh⁡(py)+Bsinh⁡(py)Where A,B are constants.Y(0)=A=0, Y(a)=Bsinh⁡(pa)=0Since p≠0, A=B=0.Case-2: λ=0, Y″=0⇒ Y(y)=A+ByY(0)=A=0, Y(a)=A+aB=0⇒ B=0Case-3: λ>0. λ=p2, p≠0Y″=−p2Y⇒ Y(y)=Acos⁡(py)+Bsin⁡(py)Using the conditions,Y(0)=A=0Y(a)=Bsin⁡(ap)=0The eigenfunctions are the nonzero solution, so to get nonzero solution, sin⁡(ap)=0⇒ p=nπa where n=1,2,...Hence eigenvalues are λn=n2π2a2 and the corresponding eigenfunctions are Yn(y)=sin⁡(nπy/a)In this case X″=n2π2a2X⇒X(x)=Cncosh⁡(nπx/a)+Dnsinh⁡(nπx/a)By Superposition principle,V(x,y)=∑n=1∞[Cncosh⁡(nπx/a)+Dnsinh⁡(nπx/a)]sin⁡(nπy/a)Using the conditions on x.V(0,y)=∑n=1∞Cnsin⁡(nπy/a)=V0Hence Cn=2a∫0aV0sin⁡(nπy/a)dy=2V0a[−anπcos⁡(nπy/a)]0a=2V0nπ(1−(−1)n)V(b,y)=∑n=1∞[Cncosh⁡(nπb/a)+Dnsinh⁡(nπb/a)]sin⁡(nπy/a)=0Cncosh⁡(nπb/a)+Dnsinh⁡(nπb/a)=0Assuming V0 is constant.V(x,y)=∑n=1∞2V0nπsinh⁡(nπb/a)(1−(−1)n)sinh⁡(nπ(b−x)/a)sin⁡(nπy/a)