Consider a square pipe of length L, where L is much larger than the dimensions of its square cross section of size a x a. The four
walls are electrically insulated from each other. Two opposite walls

Question

Consider a square pipe of length L, where L is much larger than the dimensions of its square cross section of size a x a. The four
walls are electrically insulated from each other. Two opposite walls are metallic and grounded (V=0), while the other two opposite
walls are maintained at a potential Vo. The goal is to determine the potential inside the pipe.
(a)
Assume the pipe is oriented along the z-axis. Sketch the cross section of the pipe at z = 0. Place one corner of the square at the
origin and the opposite comer at (a,a) in the xy-plane. Label each side of the cross section with its corresponding potential, choosing
the grounded walls to be at x=0 and x=a.
(b)
Explain why the potential inside the pipe does not depend on z.
(c)
Using separation of variables, assume V(x,y)=X(x)Y(y). Derive the two ordinary differential equations that result from this assumption,
and write their general solutions.
The problem is simpler if you choose X(x) to have sinusoidal solutions.
(d)
Apply the boundary conditions for the two grounded sides and write the resulting general solution.
Your answer should involve a sum
(e)
Now suppose Vo =4sin (3ттx / a)). Apply the boundary conditions for the two

VIPSolutions · Accepted Answer

# Potential Inside a Square Pipe

## (a) Cross Section Sketch and Labeling

- **Orientation:** Pipe along the $ z $-axis.
- **Cross section at $ z= $:** Square with corners at $(,)$ and $(a,a)$.
  
**Labeling:**

| Side | Potential | Coordinates | Description |
|--------|--------------|--------------|--------------|
| Left side | $ V= $ | $ x=, \,  \le y \le a $ | Grounded wall |
| Right side | $ V= $ | $ x=a, \,  \le y \le a $ | Grounded wall |
| Bottom | $ V=V_ $ | $ y=, \,  \le x \le a $ | Potential $ V_ $ |
| Top | $ V=V_ $ | $ y=a, \,  \le x \le a $ | Potential $ V_ $ |

---

## (b) Why the Potential Does Not Depend on $ z $

- **Electrostatic boundary conditions** are uniform along the length $ L $.
- **No variation** in the $ z $-direction occurs because the boundary conditions are independent of $ z $, and the problem is symmetric along the pipe's length.
- **Result:** $ V $ depends only on $ x $ and $ y $.

---

## (c) Separation of Variables and Differential Equations

Assuming:

$$
V(x,y) = X(x)Y(y)
$$

### Derivation:

The Laplace equation in 2D:

$$
\nabla^2 V = \frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} = 
$$

Applying separation:

$$
\frac{1}{X} \frac{d^2 X}{dx^2} + \frac{1}{Y} \frac{d^2 Y}{dy^2} = 
$$

Set each term equal to a separation constant $ -\lambda^2 $:

$$
\frac{d^2 X}{dx^2} + \lambda^2 X = 
$$
$$
\frac{d^2 Y}{dy^2} - \lambda^2 Y = 
$$

### General solutions:

- For $ X(x) $:

$$
X(x) = A \sin(\lambda x) + B \cos(\lambda x)
$$

- For $ Y(y) $:

$$
Y(y) = C e^{\lambda y} + D e^{-\lambda y}
$$

---

## (d) Applying Boundary Conditions and General Solution

### Boundary conditions:

- $ V= $ at $ x= $ and $ x=a $:

$$
X() =  \Rightarrow B=
$$
$$
X(a) =  \Rightarrow A \sin(\lambda a) = 
$$

To have non-trivial solutions:

$$
\sin(\lambda a) =  \Rightarrow \lambda_n = \frac{n \pi}{a}, \quad n=1,2,3,\dots
$$

### Final form of $ X_n(x) $:

$$
X_n(x) = \sin\left(\frac{n \pi x}{a}\right)
$$

### General solution:

$$
V(x,y) = \sum_{n=1}^\infty \sin\left(\frac{n \pi x}{a}\right) \left[ A_n e^{\frac{n \pi y}{a}} + B_n e^{-\frac{n \pi y}{a}} \right]
$$

---

## (e) Applying Boundary Conditions for $ V_ $

- At $ y= $:

$$
V(x,) = V_
$$

- At $ y=a $:

$$
V(x,a) = V_
$$

Expressing $ V $:

$$
V(x,y) = \sum_{n=1}^\infty \sin\left(\frac{n \pi x}{a}\right) \left[ A_n e^{\frac{n \pi y}{a}} + B_n e^{-\frac{n \pi y}{a}} \right]
$$

### Boundary conditions:

$$
V(x,) = \sum_{n=1}^\infty \sin\left(\frac{n \pi x}{a}\right) (A_n + B_n) = V_
$$

$$
V(x,a) = \sum_{n=1}^\infty \sin\left(\frac{n \pi x}{a}\right) \left[ A_n e^{n \pi} + B_n e^{-n \pi} \right] = V_
$$

Since $ V_ $ is constant, expand in Fourier sine series:

$$
V_ = \sum_{n=1}^\infty C_n \sin\left(\frac{n \pi x}{a}\right)
$$

Matching coefficients:

$$
A_n + B_n = C_n
$$
$$
A_n e^{n \pi} + B_n e^{-n \pi} = C_n
$$

Solve for $ A_n $ and $ B_n $:

$$
A_n = \frac{C_n - B_n}{1}
$$
$$
A_n e^{n \pi} + B_n e^{-n \pi} = C_n
$$

Expressed as:

$$
A_n = \frac{C_n - B_n}{1}
$$
$$
\Rightarrow (C_n - B_n) e^{n \pi} + B_n e^{-n \pi} = C_n
$$

Rearranged to solve for $ B_n $:

$$
C_n e^{n \pi} - B_n e^{n \pi} + B_n e^{-n \pi} = C_n
$$
$$
B_n \left( e^{-n \pi} - e^{n \pi} \right) = C_n (1 - e^{n \pi})
$$

$$
B_n = \frac{C_n (1 - e^{n \pi})}{e^{-n \pi} - e^{n \pi}} = \frac{C_n (1 - e^{n \pi})}{- (e^{n \pi} - e^{-n \pi})}
$$

Similarly,

$$
A_n = C_n - B_n
$$

---

## (f) Applying $ V_ = 4 \sin \left(\frac{3 \pi x}{a}\right) $

- Only the $ n=3 $ term is non-zero:

$$
C_3 = 4, \qquad C_{n \neq 3} = 
$$

Solve for $ A_3 $ and $ B_3 $:

$$
A_3 + B_3 = 4
$$
$$
A_3 e^{3 \pi} + B_3 e^{-3 \pi} = 4
$$

From the first:

$$
A_3 = 4 - B_3
$$

Substitute into second:

$$
(4 - B_3) e^{3 \pi} + B_3 e^{-3 \pi} = 4
$$

$$
4 e^{3 \pi} - B_3 e^{3 \pi} + B_3 e^{-3 \pi} = 4
$$

$$
B_3 (e^{-3 \pi} - e^{3 \pi}) = 4 - 4 e^{3 \pi}
$$

$$
B_3 = \frac{4(1 - e^{3 \pi})}{e^{-3 \pi} - e^{3 \pi}}
$$

And:

$$
A_3 = 4 - B_3
$$

**Final potential inside the pipe:**

$$
V(x,y) = \sin\left(\frac{3 \pi x}{a}\right) \left[ A_3 e^{\frac{3 \pi y}{a}} + B_3 e^{-\frac{3 \pi y}{a}} \right]
$$

where $ A_3 $ and $ B_3 $ are as above.

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