--- title: "2D Laplace potential distribution" subject: "Electromagnetism" chapter: 2 key: "semester_3/electromagnetism:laplace-2d-potential-distribution" source: https://www.bctengineer.com/q/semester_3/electromagnetism/laplace-2d-potential-distribution site: BCT Engineering --- # 2D Laplace potential distribution Electromagnetism · Chapter 2: Electric field - Asked in: 2067 Mangsir, 8 marks - Video: [Laplace's and Poisson's Equation for Electrostatic Potential](https://www.youtube.com/watch?v=cJtHMwQ218g) — Introduction to Electromagnetism - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electromagnetism/laplace-2d-potential-distribution - Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2 ## Question Use the two-dimensional Laplace equation to determine the potential distribution for the following boundary condition: V = 0 at x = 0, V = V0 at x = a, V = 0 at y = 0 and V = 0 at y = b. ## Short answer Using separation V=X(x)Y(y) with Y ∝ sin(nπy/b) satisfies y boundaries. X ∝ sinh(nπx/b) satisfies x=0 condition. Full solution: V(x,y) = (4V₀/π) Σ_{n odd} (1/n) sinh(nπx/b)/sinh(nπa/b) sin(nπy/b). ## Answer **Problem.** Inside the rectangle 0 < x < a, 0 < y < b the region is charge-free, so the potential satisfies the 2-D Laplace equation ∂²V/∂x² + ∂²V/∂y² = 0, with V = 0 on x = 0, y = 0, y = b and V = V₀ on x = a. ```text y ↑ V=0 (y=b) b +--------------+ | | V=0 | ∇²V = 0 | V=V0 (x=0) | | (x=a) +--------------+ → x 0 V=0 (y=0) a ``` #### Step 1: Separation of variables Let V(x, y) = X(x)Y(y). Substituting and dividing by XY: ```text X''/X + Y''/Y = 0 X''/X = λ² , Y''/Y = −λ² (choose sine in y because V = 0 at y = 0 and y = b) ``` So X'' − λ²X = 0 and Y'' + λ²Y = 0, giving ```text X = A cosh λx + B sinh λx (or C e^(λx) + D e^(−λx)) Y = C cos λy + D sin λy ``` #### Step 2: Boundary conditions - V = 0 at y = 0: Y(0) = 0 gives C = 0, so Y = D sin λy. - V = 0 at y = b: sin λb = 0, so λ = nπ/b, n = 1, 2, 3, … - V = 0 at x = 0: X(0) = 0 gives A = 0, so X = B sinh(nπx/b). The general solution is the sum of all modes: ```text V(x,y) = Σ(n=1→∞) c_n sinh(nπx/b) sin(nπy/b) ``` #### Step 3: Fourier coefficients using V = V₀ at x = a ```text V₀ = Σ c_n sinh(nπa/b) sin(nπy/b) for 0 < y < b ``` Multiply by sin(mπy/b) and integrate over 0 to b (orthogonality): ```text c_n sinh(nπa/b) · (b/2) = V₀ ∫₀ᵇ sin(nπy/b) dy = V₀ (b/nπ)(1 − cos nπ) c_n sinh(nπa/b) = (2V₀/nπ)(1 − cos nπ) = 4V₀/(nπ) for n odd = 0 for n even ``` #### Step 4: Final solution ```text V(x,y) = (4V₀/π) Σ(n = 1,3,5,…) (1/n) · [sinh(nπx/b) / sinh(nπa/b)] · sin(nπy/b) ``` **Check:** at x = a the ratio of sinh terms is 1, and the series (4V₀/π)Σ sin(nπy/b)/n for odd n equals V₀ for 0 < y < b. At x = 0 the sinh is zero, and at y = 0, b the sine is zero, so all other boundary conditions hold. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Substituting V = X(x)Y(y) into Laplace's equation gives which separated form? A. X''/X + Y''/Y = 0, so each ratio equals a constant of opposite sign B. X'Y' = 0 C. X''Y = Y''X = 1 D. X/X'' = Y/Y'' Answer: A. Dividing ∇²V = 0 by XY gives X''/X = −Y''/Y, so each side equals a constant (λ² and −λ²). (Key point: Separation of variables) 2. Why is Y chosen as sin(λy) and λ = nπ/b? A. V = 0 at y = 0 and y = b requires Y(0) = Y(b) = 0 B. V = V0 at y = b requires it C. Because sinh cannot be used in y D. To make the solution grow with x Answer: A. Zero potential at both y = 0 and y = b forces C = 0 and sin λb = 0, so λ = nπ/b. (Key point: Why sine in y) 3. V = 0 at x = 0 leads to which x-dependence? A. cosh(nπx/b) B. sinh(nπx/b) C. e^(nπx/b) only D. sin(nπx/b) Answer: B. X(0) = 0 removes the cosh term (A = 0), leaving X = B sinh(nπx/b). (Key point: x-dependence) 4. Applying V = V0 at x = a with orthogonality, which terms survive? A. Only even n B. All n equally C. Only odd n, with c_n sinh(nπa/b) = 4V0/(nπ) D. Only n = 1 Answer: C. (1 − cos nπ) is 2 for odd n and 0 for even n, giving 4V0/(nπ) for odd n and zero otherwise. 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