A parallel plate capacitor has a charge Q, and a voltage difference AV. The gap is air-filled with an electric field E.. 1. The capacitor is then disconnected from the circuit so that there is no path for the electrons to move from one plate to another. An insulator of dielectric constant *> 1 is then inserted to fill the entire gap. The electric field in the gap will cause the charges in the dielectric to separate slightly. This leaves an excess layer of negative charges on the top of the insulator and an excess layer of positive charges at the bottom surface of the insulator. This induced surface charge creates an induced electric field opposite the direction of the original electric field. a. The new net electric field is Ē' = Z₂ + Zinduced. Which is true of E', the magnitude of the new net field? Choose one. E' = E, (the net electric field is the same as the original electric field) E' = E₁/x (the net electric field is smaller than the original field.) b. Which is true of the new voltage difference AV* across the plates? • AV' = AV, AV' = KAV, AV/K +Q+++++ E₂ ↓ -Q₂ • C² = C₂ • E' = KE₁ (since * > 1, this implies the net electric field is larger than the original electric field). +Q++++++++++ -Qo ++++ E₂ • Bind↑ E'↓ AV¹ + + + + + + ● AV' = = c. Since the capacitor has been disconnected from the rest of the circuit, the charge cannot change: Q Q. What is the true of the capacitance of the dielectric filled capacitor compared to Co, the original capacitance? Để kho • C' = Co/K AV₂

A parallel plate capacitor has a charge Q, and a voltage difference AV. The gap is air-filled with an electric field E.. 1. The capacitor is then disconnected from the circuit so that there is no path for the electrons to move from one plate to another. An insulator of dielectric constant > 1 is then inserted to fill the entire gap. The electric field in the gap will cause the charges in the dielectric to separate slightly. This leaves an excess layer of negative charges on the top of the insulator and an excess layer of positive charges at the bottom surface of the insulator. This induced surface charge creates an induced electric field opposite the direction of the original electric field. a. The new net electric field is Ē' = Z₂ + Zinduced. Which is true of E', the magnitude of the new net field? Choose one. E' = E, (the net electric field is the same as the original electric field) E' = E₁/x (the net electric field is smaller than the original field.) b. Which is true of the new voltage difference AV across the plates? • AV' = AV, AV' = KAV, AV/K +Q+++++ E₂ ↓ -Q₂ • C² = C₂ • E' = KE₁ (since * > 1, this implies the net electric field is larger than the original electric field). +Q++++++++++ -Qo ++++ E₂ • Bind↑ E'↓ AV¹ + + + + + + ● AV' = = c. Since the capacitor has been disconnected from the rest of the circuit, the charge cannot change: Q Q. What is the true of the capacitance of the dielectric filled capacitor compared to Co, the original capacitance? Để kho • C' = Co/K AV₂

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