The capacitor is a parallel plate capacitor with a vacuum between the square plates. What is the length of a side if the distance between the plates is 0.250 mm?There is an input box for entering the value of \( s \), which represents the side length, in meters. Above the input box, there are buttons for formatting, undo, redo, and accessing a keypad. A switch that connects a battery to a 0.340 μF capacitor is closed. Several seconds later you find that the capacitor plates are charged to 30.8 μC.

The expression for the capacitance of parallel-plate capacitor is as follows: C=Aε0dA=Cdε0…

Answered: The capacitor is a parallel plate…

Similar questions

What is the capacitance of a parallel plate capacitor in vacuum? Sol. The basic definition for capacitance is: C = Since the electric potential can also be written as V = Ed Where, E is the electric field contribution of the parallel plate (two plane charge): = ol E0 %3D Sigma is the surface charge density given by: O= By, substitution, we get the capacitance of the parallel plate to be:
A parallel-plate capacitor is constructed by filling the space between two square plates with a block of three dielectric materials, as shown in the figure below. You may assume that ℓ ≫ d. A parallel-plate capacitor is comprised of two square plates with sides of length ℓ and separation distance d. There are three dielectric materials in the space between the capacitor plates. A material with length ℓ⁄2, height d, and dielectric constant ?1 fills the left half of the space. A material with length ℓ⁄2, height d⁄2, and dielectric constant ?2 fills the top right quarter of the space. A material with length ℓ⁄2, height d⁄2, and dielectric constant ?3 fills the bottom right quarter of the space. (a) Find an expression for the capacitance of the device in terms of the plate area A and d, ?1, ?2, and ?3. (Use the following as necessary: ?1, ?2, ?3, ?0, A and d.)C = (b) Calculate the capacitance using the values A = 1.30 cm2, d = 2.00 mm, ?1 = 4.90, ?2 = 5.60, and ?3 = 2.50.
Consider a parallel-plate capacitor with some non-zero charge on the plates. If we double the charge surface density on each of the plates and double the separation between the plates, how does the energy density of the capacitor change? A. The energy density stays the same. B. The energy density is halved. C. The energy density decreases by a factor of 2–√. D. The energy density increases by a factor of 4. E. The energy density increases by a factor of 2–√.
two parallel-plate capacitors (with air between the plates) are connected to a battery. Capacitor 1 has a plate area of 2.4 cm2 and an electric field (between its plates) of magnitude 1800 V/m. Capacitor 2 has a plate area of 0.56 cm2 and an electric field of magnitude 1700 V/m. What is the total charge on the two capacitors?
A parallel-plate capacitor has square plates of edge length 0.9 cm and a plate spacing of 0.029 mm. If the gap between the plates is a vacuum, what is the capacitance? F
You have a capacitor made of two sheets of foil of area 2.7 cm2 separated by a distance 0.5 mm. a) what is the capacitance (in pF(10-12F)) of this capacitor? b) what is the voltage across the capacitor if 7 x 10-12 C is stored? c) how much energy is stored in the capacitor?
b. Capacitor A and Capacitor B are two parallel plate capacitors. Capacitor B has twice the area and four times distance between plates of Capacitor A. (i) What is the ratio CB: CA, and (ii) What is the ratio of Energy stored UB:UA when the capacitors store the same charge on its plates?
A parallel plate capacitor has capacitance of 230 pF. If the area of the each plate is 60 cm2 and there is a dielectric of κ = 5 between the plates, what is the distance (in mm) between the plates?
Area = A dielectric capacitor plates A parallel plate capacitor has each plate area as 3 cm x 4 cm, separated by a 1.0 mm thick cardboard represented by the purple layer in the figure (dielectric constant € = 3.8), is fully charged. Find the capacitance (C) and the maximum charge (Qmax) on the capacitor if it's charged with an electric field E = 16 x 106 V/m.,
The dielectric of a parallel plate capacitor is to be constructed from glass (Kd= 4.6). The area of each plate is 0.50 m2 . A) What thickness should the glass have if the capacitance is to be 0.10 microfarad? B) What is the charge on the capacitor if connected to a 12-volt battery?
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? Please prove your answer. Note: This question is under the topic "Capacitance". Thank you!
1. A parallel plate capacitor is separated by a paraffined paper 0.001 cm thick (Z₁ = 2.5). The effective size of the plate is 15 by 30 cm. What is the capacitance of this capacitor?