A parallel plate capacitor, in which the space between the plates is filled with a dielectric material with dielectric constant K = 3, has a capacitance of C=870 µF and it is connected to a battery whose voltage is V=39 V and fully charged. Once it is fully charged, it is disconnected from the battery and without affecting the charge on the plates, dielectric material is removed from the capacitor. How much change occurs in the energy of the capacitor; final energy minus initial energy? Express your answer in units of mj (milli Joules) using zero decimal places.

A parallel plate capacitor, in which the space between the plates is filled with a dielectric material with dielectric constant K = 3, has a capacitance of C=870 µF and it is connected to a battery whose voltage is V=39 V and fully charged. Once it is fully charged, it is disconnected from the battery and without affecting the charge on the plates, dielectric material is removed from the capacitor. How much change occurs in the energy of the capacitor; final energy minus initial energy? Express your answer in units of mj (milli Joules) using zero decimal places.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

A parallel-plate capacitor has a capacitance of 10 mF and is charged with a 20 V power supply. The power supply is then removed and a dielectric of dielectric constant 4 is used to fill the space between the plates. The voltage now across the capacitor is O 5 V. O 80 V. O zero. O 20 V. О 10 V.
A parallel plate capacitor of capacitance C has plates of area A with separation d between them. When it is connected to a battery of voltage V, it has a charge of magnitude Q on its plates. While it is connected to the battery, the space between the plates is filled with a material of dielectric constant 3. After the dielectric is added, the magnitude of the charge on the plates and the new capacitance are: a)3Q, C b)Q, C c)Q, C/3 d)Q/3, C/3 e)3Q, 3C
A parallel plate capacitor of capacitance Co has plates of area A with separation d between them. When it is connected to a battery of voltage Vo, it has charge of magnitude Qo on its plates. While it is connected to the battery the space between the plates is filled with a material of dielectric constant 3. After the dielectric is added, the magnitude of the charge on the plates and the potential difference between them are:
A 15 µF capacitor is connected to a 50V battery and becomes fully charged. The battery is removed and a slab of dielectric that completely fills the space between the plates is inserted. If the dielectric has a dielectric constant of 5.0, (a) what is the capacitance of the capacitor after the slab is inserted? (b) What is the voltage after the dielectric is inserted? (c) How much energy is stored after the dielectric is inserted?
An air-filled capacitor with C = 4.00 uF is connected to a 6.00 volt battery. While the capacitor remains connected to the battery, a dielectric with K = 2.00 is inserted, completely filling the volume between the plates. What is the energy stored in the capacitor after the dielectric has been inserted? I know the answer is 144 uJ, but am unsure of the step-by-step solution to get this number.
A parallel-plate capacitor is fully charged by a battery. While still connected to the battery, the distance between the plates increases. What happens to the charge on the capacitor? Answers: decreases increases stays the same
A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 225 V. Assume a plate separation of d = 1.47 cm and a plate area of A = 25.0 cm². When the battery is removed, the capacitor is plunged into a container of distilled water. Assume distilled water is an insulator with a dielectric constant of 80.0. (a) Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before after AV f (b) Determine the capacitance (in F) and potential difference (in V) after immersion. Cf= = = (c) Determine the change in energy (in nJ) of the capacitor. AU = nJ AV f (d) What If? Repeat parts (a) through (c) of the problem in the case that the capacitor is immersed in distilled water while still connected to the 225 V potential difference. Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before after…
A parallel plate capacitor with plate separation do is charged by connecting it to a battery. The energy stored by the capacitor is Ug. The battery is then disconnected and the plate separation is changed to d1 = 0.7do. If U1 is the energy stored by the capacitor when the plate separation is di, what is the ratio U1/Uo? O (a) 0.7 O (b) 0.837 O (c) 0.49 O (d) 2.04 O (e) 1.43 O (f) 1.2
A 12.5 mF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 V across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them. (a) How much energy is stored in the capacitor before and after the dielectric is inserted? (b) By how much did the energy change during the insertion? Did it increase or decrease?
An engineer working in an electronics lab connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 230 V. Assume a plate separation of d = 1.38 cm and a plate area of A - 25.0 cm. When the battery is removed, the capacitor is plunged into a container of distilled water. Assume distilled water is an insulator with a dielectric constant of 80.0. d into (a) Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) O - 369 Q, = 369 before V pc after v pC (b) Determine the capacitance (in F) and potential difference (in V) after immersion. C, - 128 You will naed to include a dielectric constant when you calculate the final capacitance. Nota that x= 80 for distillad water. F AV, = (c) Determine the change in energy (in n)) of the capacitor. Au: (d) what If? Repeat parts (a) through (c) of the problem in the case that the capacitor is immersed in distilled water while still connected to the…
An air filled capactior with C= 4.0 uF is connected to a 12V battery. After the capacitor is fully charged, it is disconnected from the battery and its dielectric was changed to k =4.0 . After the new dielectric insertion, what is the new capacitance and how much energy is stored in the capacitor?
A 37 uF capacitor is being charged by a voltage source and the energy stored is equal to 249.48 uJ. While the capacitor is still connected to the voltage source, a dielectric material is inserted between the plates. As a result, the energy stored increases by 347.73 uJ. The dielectric constant of this material is equal to