Related questions
Question
Transcribed Image Text:The space between two concentric conducting spherical shells of radii b= 2.00 cm and a = 1.40 cm is filled with
a substance of dielectric constant K = 14.9. A potential difference V = 81.0 V is applied across the inner and
outer shells. Determine (a) the capacitance of the device, (b) the free charge q on the inner shell, and (c) the
charge q induced along the surface of the inner shell.
(a) Number i
(b) Number i
(c) Number
Units
Units
Units
O
O
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution
Trending nowThis is a popular solution!
Step by stepSolved in 3 steps with 6 images
Knowledge Booster
Similar questions
- Two conducting concentric spherical shells have radii a = 0.165 m and b = 0.24 m. Part (a) Express the capacitance of the two concentric shells in terms of radii a and b and the Coulomb constant k. Part (b) Calculate the numerical value of the capacitance in F. Part (c) Express the capacitance C through the potential difference△V across the capacitor and charge Q. Part (d) If the charge in the inner sphere is +Q = 3 µC, the outer sphere -Q = -3 µC, calculate the electric potential difference△V between the outside and the inside conductors in V.arrow_forwardA large capacitance of 1.58 mF is needed for a certain application. (a) Calculate the area the parallel plates of such a capacitor must have if they are separated by 4.19 µm of Teflon, which has a dielectric constant of 2.1. (b) What is the maximum voltage that can be applied if the dielectric strength for Teflon is 60 ✕ 106 V/m (c) Find the maximum charge that can be stored. (d) Calculate the volume of Teflon alone in the capacitor. (units: m^3)arrow_forwardA constant potential difference of 20.0 V is maintained between the terminals of a 5.00 µF, parallel plate air capacitor. A sheet of glass having dielectric constant K = 5.00 is then inserted between the plates of the capacitor, completely filling the space between the plates, while the battery remains connected. What is the charge induced on either face of the glass sheet (in µC)?arrow_forward
- A parallel-plate capacitor has a capacitance of 570 pF, a plate area of 400 cm², and a mica dielectric (K = 5.40) completely filling the space between the plates. At 53.0 V potential difference, calculate (a) the electric field magnitude E in the mica, (b) the magnitude of the free charge on the plates, and (c) the magnitude of the induced surface charge on the mica. (a) Number (b) Number i M. (c) Number i jeda Units Units Unitsarrow_forwardA point particle with charge q = -200 nC and mass m = 3.0x10-6 g is released from rest at the midpoint of a parallel-plate capacitor. The capacitor has plate separation d = 3.0 mm. The negative electrode is at x = 0 and the positive electrode is at x = d. The voltage across the capcitor has been measured to be AVC = 2.5 V. In this problem, ignore the weight of the particle. Give all numerical answers to two significant figures. a) What direction does the particle move? b) What is the electric field strength inside the capacitor, in V/m?arrow_forwardA parallel plate capacitor is composed of two square plates, 8.00 cm on a side, separated by a layer of dielectric of thickness 0.600 mm. The dielectric constant of the dielectric material is κ. a)If the capacitance of the capacitor is C = 2.83 × 10−10 F, then what is κ? (b) If the charge held on the capacitor’s plates is Q = 5.66 nC, then what is the potential difference between the plates? (c) What is the energy stored in the capacitor?arrow_forward
arrow_back_ios
arrow_forward_ios