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Problem 6: Two spherical conductors of radii a and b (b > a). The sphere of radius r = a is held at a potential f(0), and the sphere of radius r = b is held at a potential g(0). The problem of the potential between the two spheres, denoted by u(r, 0), can be described by 1 0²u 2 ди + r dr cot 0 du + r2 d0 = 0, a

Problem 6: Two spherical conductors of radii a and b (b > a). The sphere of radius r = a is held at a potential f(0), and the sphere of radius r = b is held at a potential g(0). The problem of the potential between the two spheres, denoted by u(r, 0), can be described by 1 0²u 2 ди + r dr cot 0 du + r2 d0 = 0, a

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Problem 6: Two spherical conductors of radii a and b (b > a). The sphere of radius r = a is held at a potential f(0), and the sphere of radius r = b is held at a potential g(0). The problem of the potential between the two spheres, denoted by u(r, 0), can be described by 2 ди 1 0Pu cot 0 du + 0, a <r < b, 0 < 0 < T, dr2 r dr p2 002 r2 d0 u(r, 0) is finite, u(r, 7) is finite, u(а, 0) — f(0), u(b, 0) = g(0). Solve it.
Transcribed Image Text:Problem 6: Two spherical conductors of radii a and b (b > a). The sphere of radius r = a is held at a potential f(0), and the sphere of radius r = b is held at a potential g(0). The problem of the potential between the two spheres, denoted by u(r, 0), can be described by 2 ди 1 0Pu cot 0 du + 0, a <r < b, 0 < 0 < T, dr2 r dr p2 002 r2 d0 u(r, 0) is finite, u(r, 7) is finite, u(а, 0) — f(0), u(b, 0) = g(0). Solve it.
Problem 3: A transistor extends over the interval r E [0, L]. The concentration of positive charge carriers is denoted by u(x,t). It can be shown that u(x, t) satisfies the equation Urr aUr = Ut, 0 < x < L, t > 0, u(0, t) = 0, u(L, t) = 0 u(x, 0) = f(x) 1 where a is a positive constant. Solve it.
Transcribed Image Text:Problem 3: A transistor extends over the interval r E [0, L]. The concentration of positive charge carriers is denoted by u(x,t). It can be shown that u(x, t) satisfies the equation Urr aUr = Ut, 0 < x < L, t > 0, u(0, t) = 0, u(L, t) = 0 u(x, 0) = f(x) 1 where a is a positive constant. Solve it.
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