A capacitor consists of two concentric conducting spherical surfaces of radii a and b (a < b)and charge of Q. The medium between the conductors has a dielectric constant (relative1permittivity) of ɛ, = where R is the radial distance.Ra) Using Gauss' law find the electric field intensity Ē between the plates.b) Find the electric potential difference between the conductors.c) What is the capacitance?d) Find the energy stored in the capacitor in two ways (your results should match):• Using capacitance found in part (c) based on W. = Q²/C.21Using the energy density we = ɛE? and integrating over the entire volume ofthe capacitor.2

Answered: Image /qna-images/answer/6b9b4e10-becf-4921-b3b2-63bb039d1f0b.jpg

Answered: A capacitor consists of two concentric…

Similar questions

Consider a parallel-plate capacitor made up of two conducting plates with dimensions 34 mm × 18 mm. a) If the separation between the plates is 1.1 mm, what is the capacitance, in pF, between them? b) If there is 0.32 nC of charged stored on the positive plate, what is the potential, in volts, across the capacitor? c) What is the magnitude of the electric field, in newtons per coulomb, inside this capacitor? d) If the separation between the plates doubles, what will the electric field be if the charge is kept constant?
Imagine a parallel plate capacitor with mica between the plates. Suppose the potential difference between the plates is 82V, and there is an electric field between the plates with a strength of 320 N/C. If each plate has an area of 0.030 square meters, a) What is the separation between the plates? b) Find the capacitance of the capacitor. c) How much charge is on each plate? d) How much energy is stored in the capacitor? For each part, explain your reasoning.
A simple capacitor can be constructed from two conductive plates. Two conductive plates with an area of 10.0cm x 10.0 cm are held facing one another at a separation of 1.50 mm and 12.0 V is applied between them. a) Find the capacitance of this configuration b) Find the charge on the plates. c) How much energy is stored in the capacitor? d) How much energy can the capacitor store if a dielectric with K= 4.5 is inserted between the plates rather than air?
Kto A In the formula for capacitance C-- variable d represents: Select one: O separation between the plates of the capacitor O area of the metal plates that make up the capacitor for this experiment. O is the permittivity constant. O is the dielectric constant of the insulating material between the plates. When two capacitor plates are connected to a potential difference V, Select one: O None of the above One of the plates is charged with charge Q and the other plate is neutral. O The two plates are charged with equal but opposite amount of charge Q O The two plates are charged with equal amount of charge Q
When a battery is connected across an empty capacitor, the charge on its plates is Q0 (fig. a). If a dielectric of dielectric constant ?κ is inserted between the plates while the battery remains in place (fig. b), what is the expression for the induced charge on the dielectric surface in terms of the original plate charge Q0?
A parallel-plate capacitor filled with air has plates of area 6.5×10−3 m2 and a separation of 0.32 mm. a) Find the magnitude of the charge on each plate when the capacitor is connected to a 11-V battery. (Q = ? nC) b) Why does answer A decrease if the seperation between the plates increases? c) Calculate the magnitude of the charge on the plates if the separation is 0.90 mm. (Q = ? nC)
magnitude and the direction of the electric field between the plates? Two capacitors give an equivalent capacitance of C, when connected in parallel and an tyuivalent capacitance of C, when connected in series. What is the capacitance of each сaрacitor?