Find the charge on each of the capacitors in the figure below.
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A: QT=Q1+Q2Q=CVQ1=C1VQ2=C2V
![Find the charge on each of the capacitors in the figure below.
1.00 µF.
C5.00 µF
24.0 V
8.00 µF
4.00 µF
1.0-uF capacitor
5.0-µF capacitor
8.0-µF capacitor
4.0-µF capacitor](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0934b2e8-bd06-4f5a-8888-c940a894eb36%2Fc7c5aa46-b3d5-48e9-bda3-0c4c236ece00%2Fxx0ijd_processed.jpeg&w=3840&q=75)
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- The voltage across an air-filled parallel-plate capacitor is measured to be V, =194.0 V. When a dielectric is inserted and completely fills the space between the plates, as in the figure below, the voltage drops to V2 -57.1 V. Dielectric AV AV da) What is the dielectric constant of the inserted material? Can you identify the dielectric? O bakelite O paper O neoprene rubber O nylon O teflon (b) If the dielectric doesn't completely fill the space between the plates, what could you conclude about the voltage across the plates?What are the two methods (formulas) to find the energy density of a capacitor? can you please write neatly and expain each variableTwo identical capacitors (A and B) with capacitance Co are hooked up to a battery of voltage Vo. A dielectric of constant ko is added between the plates of capacitor B while it is connected to the battery, as in the figure at right. a) Is the charge on capacitor A (QA) greater than, less than, or equal to the charge on capacitor B (QB)? What about the voltages (VA and VB)? Explain your reasoning. b) Vo CB The capacitors above are carefully disconnected from the battery (without being discharged), after the battery is disconnected the dielectric slab is removed from capacitor B. The capacitors remain connected to each other the entire time. Determine an expression for the final charges on each of the capacitors (QA₁f and QB₁f) after the dielectric has been removed in terms of the given constants (Ko, Co, Vo). Show your work and explain your reasoning.