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Finite Parallel Plate Capacitor

Finite Parallel Plate Capacitor. Parallel Plate Capacitor. C = ² 0 A/d. Area of plates = A ( m 2 ) Plate separation = d (m). ¢ V = Q / C. -Q. +Q. C = K ² 0 A/d. ¢ V’ = ¢ V / K. -Q. +Q. In general, for ANY capacitor, dielectric: Reduces ¢ V by factor K for same Q

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Finite Parallel Plate Capacitor

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  1. Finite Parallel Plate Capacitor Brian Meadows, U. Cincinnati

  2. Parallel Plate Capacitor C = ²0 A/d Area of plates = A (m2) Plate separation = d (m) ¢V = Q / C -Q +Q C = K²0 A/d ¢V’ = ¢V / K -Q +Q • In general, for ANY capacitor, dielectric: • Reduces ¢V by factor K for same Q • I.e. it Increases C by factor K Brian Meadows, U. Cincinnati

  3. Capacitor with Partial Dielectric C (e.g., if parallel plate C= ²0 A/d) -Q/2 +Q/2 ¢V = Q / C -Q/2 +Q/2 C = ?? -Q/2 +Q/2 To keep ¢V the same We need to increase the charges: -KQ/2 +KQ/2 ¢V = (Q/2+KQ/2)/C’ C’  C(1+K)/2 Brian Meadows, U. Cincinnati

  4. Capacitors in Parallel -Q1 +Q1 Q = Q1+Q2 ¢ V = Q/(C1+C2) So C’ = C1+C2 C1 -Q2 +Q2 C2 Brian Meadows, U. Cincinnati

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