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Total energy of charge in parallel plate capacitor
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[QUOTE="Inquisitive Student, post: 5892572, member: 635921"] From boundary conditions we know that Φ(x = 0) = 0 (stated in the problem that left wall of parallel plate capacitor is at Φ = 0) and that Φ(x = L) = V. Poisson's equation in space (we neglect the charge in the problem, we are just finding the potential due to the walls) is ∇[SUP]2[/SUP]Φ = 0, or in 1D: $$\frac{\partial^2 Φ}{ \partial x^2} = 0$$. Since the second derivative is equal to 0, that means the first derivative was equal to a constant. So, $$\frac{\partial Φ}{\partial x} = c$$, where c is a constant. We can solve this differential equation by integrating, $$\frac{\partial Φ}{\partial x} = c$$ becomes ∫∂Φ = ∫c ∂x, which means $$Φ(x) = cx + c_1$$ where c[SUB]1[/SUB] is a constant that came from integration. Now plugging in the boundary conditions: Φ(0) = 0 gives c[SUB]1[/SUB] = 0 Φ(L) = V gives v = c*L or c = v/L So in total we get: Φ(x) = cx = vx/L [/QUOTE]
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Total energy of charge in parallel plate capacitor
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