5. Consider two situations Case W: Two conducting parallel plates with area A and separated by a distance d form a capacitor, they are charged with a potential difference of and left connected to the battery. Case X: The same conducting parallel plates in case W with area A and separated by a distance d with a neoprene dielectric in between form a capacitor, they are charged with a potential difference of and left connected to the battery. A = 9 x 10² m² d = 3 x 102 m = 9V Fill in the following table with values for both Case W and Case X Capacitance Charge on Capacitor Potential Difference across Capacitor (Uniform) Electric Field magnitude in the middle of the capacitor Case W Case X

5. Consider two situations Case W: Two conducting parallel plates with area A and separated by a distance d form a capacitor, they are charged with a potential difference of and left connected to the battery. Case X: The same conducting parallel plates in case W with area A and separated by a distance d with a neoprene dielectric in between form a capacitor, they are charged with a potential difference of and left connected to the battery. A = 9 x 10² m² d = 3 x 102 m = 9V Fill in the following table with values for both Case W and Case X Capacitance Charge on Capacitor Potential Difference across Capacitor (Uniform) Electric Field magnitude in the middle of the capacitor Case W Case X

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 is charged by being connected to a battery and is kept connected to the battery. The separation between the plates is then doubled. Which of the following is true? Select one. a. D increases, E increases b. D remains the same, E increases c. D decreases, E increases d. D increases, E remains the same e. D remains the same, E remains the same f. D decreases, E remains the same g. D increases, E decreases h. D remains the same, E decreasesi. D decreases, E decreases
17. A 40 mF parallel plate capacitor is connected to a 12.0 V battery as shown in the figure, thus charging the plates which are 1.0 mm apart. (a) What is the net charge of the right plate of the capacitor?
Increasing the distance between the plates of an isolated charged capacitor has what effect? A) It increases the electric field between the plates. B) It decreases the energy stored in the capacitor. C) It increases the charge on the plates. D) It increases the voltage difference between the plates
A parallel-plate capacitor is fully charged by a battery. Now the battery is disconnected, and the distance between the plates increases. What happens to the electric field in the capacitor? Answers: stays the same increases decreases
Question 17 ( The effective area of each plate of a parallel plate capacitor is 2.1 m2. The capacitor is filled with neoprene rubber (k = 6.4). When a 6.0-V potential difference exists across the plates of the capacitor, the capacitor stores 4.0 µC of charge. a) What is the capacitance of the capacitor? (-. b) Determine the plate separation of the capacitor.(
Question 13 The plates of a 1.50 pF parallel plate capacitor are separated by 1.00x10-4 m. The space between the plates is filled with a material with a dielectric constant of 2.26. What is the area of the plates? 1.00x10-6 m² 4.35x10-6 m² 7.50x10-6 m² 9.00x10-6 m²
Capacitance Problem 3: Two parallel plates have charges +Q and −Q, as shown. The magnitude of the charges is 29.7nC, the area of each plate is A=0.0105 m2, and the distance between them is d = 8.53cm. The plates are surrounded by air. Part (f) If the area is doubled to 2A while the gap is decreased to 14d, then what is the value, in picofarads, of the new capacitance? Part (g) What is the new value, in volts, of potential difference between the plates with the original charges, +Q and −Q, given the increased area and reduced gap of Part (f)?
25. You have two flat metal plates each of area 1.00 m², with which to construct a parallel- plate capacitor. A) If the capacitor of the device is to be 1.00 F, what must be the separation of the plates? B) Could the capacitor actually be constructed?
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)
There are plates between a parallel plates capacitor. They are seperated by distance of 4.2 mm with each having an area of 9.75 cm2. We know that each plate has charge 45 pC. Looking at a x and y coordinate system, the charge on q1 = 6x10^-9 C is on ( 0 , 0.3m ) and q2 charge = -4x10^-9 C is placed ( 0 , -0.3m ) a. Find electric potential at origin based on the two charges (volts please) b. Find electrical potential (Vp) at point P on (0.4 m , 0 ) c. Find total word done in micro joules if third point charge (Q = 5x10^-6 C) is moved from point ) to point P
An air capacitor is made up of two parallel plates, each with an area of 10.7 cm^2 separated at a distance of 2.5 mm. A potential difference of 44 volts is applied to these plates. Calculate the surface charge density. answer nC/m^2 and two decimal places