| Author | Message | |||
     Devesh Khanal (dr_chiao) Member Username: dr_chiao Post Number: 7 Registered: 05-2007 |
Mr Nelson, In metallic electrostatics, after I solve the equation, How do I calculate the total surface charge? To be more specific, suppose I have an parallel-plate capacitor, and the Laplace equation has been solved with FlexPDE. I want the total surface charge on the surface of one of the plates. I tried to use Q=SURF_INTEGRAL(Normal(Grad(phi))*epsilon0,"plateA-surface"), but the result is much much smaller than C*V, where C was calculated from analytical expression. Any suggestions? Thanks. | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 863 Registered: 06-2003 |
Are you sure you integrated over an entire closed surface? You might have to sum more than one integral. | |||
| Devesh Khanal (dr_chiao) Member Username: dr_chiao Post Number: 8 Registered: 05-2007 |
Here I attached my code, to calculate a cylinder-plane system, where the cylinder has finite length and is grounded, and the plane is at voltage "Vgate". I tried to calculate capacitance from two different methods: from energy: C*Vgate^2/2=Integral(eps*(grad(V))^2), or from charge: C=Sintegral(normal(eps*grad(V)))/Vgate but they do not agree  please help. Thanks. | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 870 Registered: 06-2003 |
I don't know what is happening here. I created a 3D version of our Application Manual test problem Capacitance.pde, and the two computations agree to 3 percent (see attached). You have a sharp edge at the top of the can, at which point the E field is in principle infinite. It is possible that the strong variations in E over a few cells destroys the accuracy of the numerical integration. I have not verified this conjecture.
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cylinder-plane capacitor