Please help with the 2 questions in example 26.1 As Checkpoint 26.6 demonstrates, the effect of the slabis to make the electric field in part of thecapacitor plates zero, thus reducing the magnitude of thepotential difference between the plates for a given amountof charge on them. As the next example shows, the converseof this fact is that, for a given potential difference between itsplates, the capacitor can store more energy with the slab in-serted than it can without the slab, In otheT words, the slabincreases the capacitor's capacity to store charge.spacebetween theExample 26.1 Metal-slab capacitorinSuppose the capacitor in Figure 26.9 has a plate separation dis-Lance d and the plates carry charges +q and 14 when the capaci-lor is connected to a battery that maintalns potential differenceVan between its terminals. Ifa metal slab of thickness d/2 Is in-serted midway between the platles while the batlery remains con-nected, what happens to (a) the magnitude of the electric fieldbetween the plales and (b) the quantity of charge on the plates?Outarthed/the(b)douGETTING STARTED I am given the plate separalion distanceand plate charges for a capacitor connected to a battery and Imust delermine how the electric field magnitude between theplates and the quantily of charge on each plate change when ametal slab is inserted. Because the capacitor remains connectedto The battery, the potential difference across the capacitor isUxed. Because the slab is conducting, the electric field inside theab is always ZETO.mea.conntweerIknoCnce.decrea Figure 26.9 Inserting a conducting slab between the plates of a parallel-plate capacitor(a)(b)meta sरab+4Teetren altengg themseiveN Ao thatlecticficl wthin sla N zero.26.6 Suppose flie capacilor n Tigure 26.918 charged andthen dsouecled from the battery. () She cond ing slabs inserted in the capacitor, as in Figure 26.9 does the amountDl tharge onfhe capacitor plates incease, decredse. or slay thcsanie )Hopuch charge accunulate hside of theslab onte itAnsurted? a What is the potential ditlerenceacross thhe nuela

When you insert a metal slab or a conducting slab inside a parallel plate capacitor, its capacitance…

Answered: Suppose the capacitor in Figure 26.9…

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter20: Electric Potential And Capacitance

Section: Chapter Questions

Problem 5CQ

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I am unsure of where I went wrong in my calculations. If possible, please list the steps so I may understand where I went wrong. (a). Ho wmuch charge does a battery have to supply a 5.00 uF capacitor to create a potential difference of 1.20 V across it's plates? (in C) (b). How much energy is stored in the capacitor in this case? (in J) (c). How much charge would the battery have to supply to store 1.50 J of energy in the capacitor? (in C) (d). What would be the potential across the capacitor in that case? (in V)
Please provide a drawing for the problem and solve, thank you. A capacitor of 3.23µF has an area of 6.35mm^2. Determine the separation distance between the two plates. If the new capacitance is now 5.63mF, what is the value of the dielectric inserted to the capacitor? What is the new voltage if the capacitor is connected to a 110 volts source? If the electric field created by two plates 8.99x10^4 N/C and a working voltage of 63 volts. Will the capacitor experience a dielectric breakdown? Prove your answer.
Please write out the solution,that way I can follow along and understand the steps, if it's in the "typed out" format is confusing to follow the thought process. Thank you so much!!
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First part of the question: Two identical capacitors are connected in parallel and charged with a 20-V power supply as shown in the figure on the left. After the charges on the capacitors have reached their final values, the charge on each capacitor is 10 uC. Then the power supply is disconnected and a dielectric with K = 4 is inserted to fill the space !3D between the plates of one of the capacitors, a) Find now the charge on the capacitor with dielectric. Give your answer in microcoulombs.
In the Figure 2 above, all plates are square and have same area. Left capacitor called Cair and its value is known as 2mF. Right capacitor has dielectric constant K=1.5 in d distance. Other d distance it is also covered with air. Potential difference between A and B VAB= 10 V. According to all these given information; (a) Write and find equivalent capacitance of the system. (b) What is the total energy stored in these capacitors. ( Hint: Energy stored in equivalent capacitor)
In the Figure 2 above, all plates are square and have same area. Left capacitor called Cair and its value is known as 2mF. Right capacitor has dielectric constant K=1.5 in d distance.Other d distance it is also covered with air. Potential difference between A and B VAB= 10 V. According to all these given information; (a) Write and find equivalent capacitance of the system. (b) What is the total energy stored in these capacitors. ( Hint: Energy stored in equivalent capacitor)
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Two dielectric slabs with dielectric constants, K, and K2 are placed in parallel plate capacitor as shown in fig. Plates of the capacitor and both slabs have the same width of 15d. If the capacitance of the capacitor before slabs are inserted is Co, What is the capacitance of capacitor with slabs in terms of Co ? Your answer: O A) 3.7Co O B) 4.8C, O C)2.1Co O D) 1.4C, O E) 5.6C,
Can you solve the question in the picture shown?
To operate a given flash lamp requires a charge of 38 mC. Whatcapacitance is needed to store this much charge in a capacitor witha potential difference between its plates of 9.0 V?

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