# can someone walk me through this --- you'd be a lifesaver

a rectangular pipe running parallel to the z axis(-�‡ to +�‡) has 3 grounded metal sides at y=0, y=a and x=0. the fourth side at x=b is maintained at a specified potential v0(y).
a. Develop a general formula for the potential within the pipe
b. Find the potential explicitly for the case v0(y)=constant
c. find the net charge per unit length on the side opposite v0

Solve Laplace's equation using separation of variables. Choose trignometric functions as a function of y and hyperbolic fnctions as a function of x.

a. To develop a general formula for the potential within the pipe, we need to solve Laplace's equation using separation of variables. In this case, we choose trigonometric functions as a function of y and hyperbolic functions as a function of x.

Let's assume the potential within the pipe can be represented as V(x, y) = X(x) * Y(y). Plugging this into Laplace's equation, we have:

(d^2V/dx^2) + (d^2V/dy^2) = 0

Expanding the derivatives and substituting V(x, y) = X(x) * Y(y), we get:

X''(x) * Y(y) + X(x) * Y''(y) = 0

Dividing by X(x) * Y(y), we separate the variables to obtain:

X''(x) / X(x) = -Y''(y) / Y(y)

Since the left side depends only on x and the right side depends only on y, both sides must be equal to a constant, which we will denote as -λ^2.

X''(x) / X(x) = - λ^2 ---> (1)
Y''(y) / Y(y) = λ^2 ---> (2)

For equation (1), assuming b > 0 (since x ranges from 0 to b), the general solution will be:

X(x) = A * cosh(λx) + B * sinh(λx)

For equation (2), assuming a > 0 (since y ranges from 0 to a), the general solution will be:

Y(y) = C * cos(λy) + D * sin(λy)

Combining the solutions, we have:

V(x, y) = (A * cosh(λx) + B * sinh(λx)) * (C * cos(λy) + D * sin(λy))

b. For the case v0(y) = constant, we can assume that V(x, y) = V_0 is the constant potential maintained at x = b. Therefore, we have:

V_0 = (A * cosh(λb) + B * sinh(λb)) * (C * cos(λy) + D * sin(λy))

To find the potentials A, B, C, and D, we can use the boundary conditions given. The three grounded metal sides at y = 0, y = a, and x = 0 suggest that the potential must be zero at those points. Therefore, we have:

V(0, y) = 0 ---> (A * cosh(0) + B * sinh(0)) * (C * cos(λy) + D * sin(λy)) = 0

This equation implies that either A or B is zero. Let's assume B = 0 and A ≠ 0.

V(b, y) = V_0 ---> (A * cosh(λb)) * (C * cos(λy) + D * sin(λy)) = V_0

This equation implies that C must be zero since cosh(λb) ≠ 0. Assuming C = 0 and D ≠ 0, we have:

V(x, 0) = 0 ---> (A * cosh(λx) + B * sinh(λx)) * (D * sin(0)) = 0

This implies that D must be zero, which contradicts our assumption.

Therefore, the potential V(x, y) = constant cannot satisfy all the given boundary conditions. The general formula for the potential within the pipe cannot be given explicitly for this case.

c. To find the net charge per unit length on the side opposite V_0, we need to determine the derivative of the potential along that side with respect to y, and multiply by the permittivity.

Taking the derivative of the potential V(x, y) = (A * cosh(λx) + B * sinh(λx)) * (C * cos(λy) + D * sin(λy)) with respect to y, we have:

dV(x, y)/dy = λ * (A * cosh(λx) + B * sinh(λx)) * (-C * sin(λy) + D * cos(λy))

Using this expression, you can evaluate the derivative at the desired location and multiply by the permittivity to obtain the net charge per unit length on the side opposite V_0.