T/2 = In(2)T.
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In the box below, show that the time it takes the voltage across a discharging capacitor to drop to one-half its initial value is also given by Eq.(5)
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- A capacitor is constructed from two square, metallic plates of sides ℓ and separation d. Charges +Q and -Q are placed on the plates, and the power supply is then removed. A material of dielectric constant κ is inserted a distance x into the capacitor as shown in the figure below. Assume d is much smaller than x. Suggestion: The system can be considered as two capacitors connected in parallel. (Use the following as necessary: ε0, κ, ℓ, Q, d, and x.) (a) Find the equivalent capacitance of the device.Ceq = (b) Calculate the energy stored in the capacitor.U =Explain why the work needed to move a charge Q through a potential difference AV is W = QAV, whereas the energy stored in a charged capacitor is U = 1/2 QAV. This answer has not been graded yet. Where does the 1/2 factor come from?For the system of capacitors shown in the figure below, find the following. (Let C, = 3.00 µF and C, = 1.00 µF.) %3D 6.00 µF 2.00 µF C, 90.0 V (a) the equivalent capacitance of the system µF (b) the charge on each capacitor on C1 µC on C2 on the 6.00 pF capacitor µC on the 2.00 pF capacitor (c) the potential difference across each capacitor across C, V across C, V across the 6.00 µF capacitor V across the 2.00 µF capacitor V