A parallel-plate capacitor is constructed using a dielectric material whose dielectric constant is 2.90 and whose dielectric strength is 2.20 x 10 V/m. The desired capacitance is 0.100 uf, an capacitor must withstand a maximum potential difference of 4.00 kV. Find the minimum area of the capacitor plates 0.39 x Find an expression for the plate separation that at the maximum potential difference just barely leads to dielectric breakdown Use that separation to find the area that would give the desires

A parallel-plate capacitor is constructed using a dielectric material whose dielectric constant is 2.90 and whose dielectric strength is 2.20 x 10 V/m. The desired capacitance is 0.100 uf, an capacitor must withstand a maximum potential difference of 4.00 kV. Find the minimum area of the capacitor plates 0.39 x Find an expression for the plate separation that at the maximum potential difference just barely leads to dielectric breakdown Use that separation to find the area that would give the desires

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 7.44 µF isolated capacitor is constructed with Teflon, having a dielectric constant of 2.10, between the plates. The capacitor is initially charged to 1.34 volts, and then the Teflon is removed. After removing the Teflon, what is the potential difference across the capacitor plates?
A parallel plate capacitor has a plates area A=0.09 m² and 18 nF capacitance is connected to a potential difference of V= 45 volt. Calculate the surface charge density o (in µC/m²) on the plate.
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 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 charge of magnitude Q on its plates. It is then disconnected from the battery and the space between the plates is filled with a material of dielectric constant 3. After the dielectric is added, the magnitudes of the capacitance and the potential difference between the plates are: 3C and V O 3C and V/3 C/3 and 3V C/3 and V
Two parallel plates having equal and opposite charges with area L x L separation d is filled with a dielectric material as shown in the Figure. The related dielectric constant varies with zask=a+ Here, a 46 cm, b -92 cm, d 0.31 cm and L=1m are given. Find the capacitance of this capacitor in 10-11 Farad. Take o=10-¹¹ C²/Nm². Answer:
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 certain substance has a dielectric constant of 2.2 and a dielectric strength of 22 MV/m. If it is used as the dielectric material in a parallel-plate capacitor, what minimum area should the plates of the capacitor have to obtain a capacitance of 10 × 10²µ F and to ensure that the capacitor will be able to withstand a potential difference of 3.4 kV? Number i Units
A parallel plate capacitor is made of dielectric material whose dielectric constant is 3.00 and whose dielectric strength is 2.00 x 108 V / m. The desired capacitance is 0.250 mF and the capacitor must withstand a maximum potential difference of 4.00 kV. Determine the minimum area of the plates of this capacitor
An air-filled parallel-plate capacitor has a capacitance of 2 pF. The plate separation is then halved and a wax dielectric is inserted, completely filling the space between the plates. As a result, the capacitance becomes 8 pF. The dielectric constant of the wax is:
A parallel-plate capacitor is constructed using a dielectric material whose dielectric constant is 3.30 and whose dielectric strength is 1.20 x 108 V/m. The desired capacitance is 0.300 μF, and the capacitor must withstand a maximum potential difference of 4.00 kV. Find the minimum area of the capacitor plates. m²
An air filled capacitor of Co= 2.0 µF is fully charged with Qo-42.0 µC using an electromotive force of ɛ, as shown in the figure. The capacitor is then disconnected from the battery and filled with a dielectric material of constant K. If the electric potential across the capacitor falls to 1.9 V, then find K. Q. K
A 10 nF air capacitor is connected to a power supply, the energy stored in the capacitor is 125 x 10 J. While the capacitor is kept connected to the power supply, a slab of dielectric is inserted that completely fills the space between the plates. This increases the stored energy by 375 x 10- J. What is the potential difference between the capacitor plates after the dielectric is inserted? 200 V 600 V 700 V 500 V 300 V