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The capacitance for a parallel plate capacitor

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The capacitance for a parallel-plate capacitor changes to: Note that the dielectric constant is a unitless variable.
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Dielectric In general, you can say that in a region completely filled by a dielectric material of dielectric constant κ , that all equations containing ε o are replaced with κε o . If you have a partially filled region, the common thing to do is turn to Gauss’ Law. For any given plate separation, there is a maximum electric field that can be produced in the dielectric before it breaks down and begins to conduct. This maximum electric field is called the dielectric strength (measured in N/C).
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Energy of a Capacitor Suppose you have a given capacitor charged to a potential difference Δ V with a charge ±q on either plate. How much energy would it take to transfer an small amount of charge dq? Start with the relationship between work and electric potential: If we wanted to calculate the entire work done to fully charge the capacitor in this manner (from q = 0 to q = Q) we have: dW = ! V ( ) dq W = dW ! = q C " # $ % & ' dq 0 Q ! dW = q C ! " # $ % & dq
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Energy of a Capacitor Since C merely depends on the geometry of the capacitor, we have: Since the electric force is conservative, the energy stored (U elec ) in the capacitor is equivalent to the work put in: Energy stored = 1 2 Q ! V ( ) = 1 2 C ! V ( ) 2 = Q 2 2 C W = 1 C q ( ) dq 0 Q ! W = 1 C Q 2 2 ! " # $ % & The main use of a capacitor is to store and then discharge energy.
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Capacitance Example A 3.55 μ F capacitor (C 1 ) is charged to a potential difference Δ V o =6.30V, using a battery. The charging battery is then removed, and the capacitor is connected to an uncharged 8.95 μ
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