Q1/ A parallel plate capacitor having a plate area of ( 20 cm') and plate separation of ( 2 mm ) and it is charged by (100 volt) battery. The battry is then removed. Calculate 1- The value of ( C) 2- The charge on the plates of the capacitor 3- The electric field between the plates 4- The energy stored in the capacitor ?

Q1/ A parallel plate capacitor having a plate area of ( 20 cm') and plate separation of ( 2 mm ) and it is charged by (100 volt) battery. The battry is then removed. Calculate 1- The value of ( C) 2- The charge on the plates of the capacitor 3- The electric field between the plates 4- The energy stored in the capacitor ?

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter8: Capacitance

Section: Chapter Questions

Problem 53P: A parallel-plate capacitor has charge of magnitude 9.00F on each plate and capacitance 3.00F when...

See similar textbooks

Similar questions

A capacitor has parallel plates of area 12 cm2 separated by 2.0 mm. The space between the plates is filled with polystyrene, (a) Find the maximum permissible voltage across the capacitor to avoid dielectric breakdown, (b) When the voltage equals the value found in part (a), find the surface charge density on the surface of the dielectric.
In different experimental trials, an electron, a proton, or a doubly charged oxygen atom (O--), is fired within a vacuum tube. The particle's trajectory carries it through a point where the electric potential is 40.0 V and then through a point at a different potential. Rank each of the following cases according to the change in kinetic energy of the particle over this part of its flight from the largest increase to the largest decrease in kinetic energy. In your ranking, display any cases of equality, (a) An electron moves from 40.0 V to 60.0 V. (b) An electron moves front 40.0 V to 20.0 V. (c) A proton moves from 40.0 V to 20.0 V'. (d) A proton moves from 40.0 V to 10.0 V. (e) An O-- ion mines from 40.0 V to 60.0 V.
When a 360-nF air capacitor is connected to a power supply, the energy stored in the capacitor is 18.5J . While the capacitor is connected to the power supply, a slab of dielectric is insetted that completely fills die space between the plates. This increases the stored energy by 23.2J . (a) What is the potential difference between the capacitor plates? (b) What is die dielectric constant of the slab?
A capacitor is constructed from two square, metallic plates of sides and separation d. Charges +Q and Q are placed on the plates, and the power supply is then removed. A material of dielectric constant is inserted a distance x into the capacitor as shown in Figure P25.49 (page 690). Assume d is much smaller than x. (a) Find the equivalent capacitance of the device. (b) Calculate the energy stored in the capacitor. (c) Find the direction and magnitude of the force exerted by the plates on the dielectric. (d) Obtain a numerical value for the force when x = /2, assuming = 5.00 cm, d = 2.0 mm, the dielectric is glass ( = 4.50), and the capacitor was charged to 2.00 103 V before the dielectric was inserted. Suggestion: The system can be considered as two capacitors connected in parallel. Figure P25.49
A parallel-plate capacitor with plates of area LW and plate separation t has the region between its plates filled with wedges of two dielectric materials as shown in Figure P25.48. Assume t is much less than both L and W. (a) Determine its capacitance. (b) Should the capacitance be the same if the labels 1 and 2 are interchanged? Demonstrate that your expression does or does not have this property. (c) Show that if 1 and 2 approach equality to a common value , your result becomes the same as the capacitance of a capacitor containing a single dielectric: C = 0LW/t. Figure P25.48
a parallel-plate capacitor with area 0.200 m2 and plate separation of 3.00 mm is connected to a 6.00-V battery. (a) What is the capacitance? (b) How much charge is stored on the plates? (c) What is the electric field between the plates? (d) Find the magnitude of the charge density on each plate. (e) Without disconnecting the battery, the plates are moved farther apart. Qualitatively, what happens to each of the previous answers?
a parallel-plate capacitor with area 0.200 m2 and plate separation of 3.00 mm is connected to a 6.00-V battery. (a) What is the capacitance? (b) How much charge is stored on the plates? (c) What is the electric field between the plates? (d) Find the magnitude of the charge density on each plate. (e) Without disconnecting the battery, the plates are moved farther apart. Qualitatively, what happens to each of the previous answers?
A parallel-plate capacitor with only air between its plates is charged by connecting the capacitor to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates, (a) A voltmeter reads 45.0 V when placed across the capacitor. When a dielectric is inserted between die plates, completely filling the space, the voltmeter reads 11.5 V. What is the dielectric constant of the material? (b) What will the voltmeter read if the dielectric is now pulled away out so it fills only one-third of the space between the plates?
A capacitor is made from two flat parallel plates placed 0.40 mm apart. When a charge of 0.020C is placed oil the plates the potential difference between them is 250 V. (a) What is the capacitance of the plates? (b) What is the area of each plate? (c) What is the charge on the plates when the potential difference between them is 500 V? (d) What maximum potential difference can be applied between the plates so that the magnitude of electrical fields between the plates does not exceed 3.0 MV/m?
An air-filled (empty) parallel-plate capacitor is made from two square plates that are 25 cm on each side and 1.0 mm apart. The capacitor is connected to a 50-V battery and fully charged. It is then disconnected from the battery and its plates are pulled apart to a separation of 2.00 mm. (a) What is the capacitance of this new capacitor? (b) What is the charge on each plate? (c) What is the electrical field between the plates?
For a Teflon ™-filled, parallel-plate capacitor, the area of the plate is 50.0 cm and the spacing between the plates is 0.50 mm. If the capacitor is connected to a 200-V battery, find (a) the free charge on the capacitor plates, (b) the electrical field in the dielectric, and (c) the induced charge on the dielectric surfaces.
A parallel-plate capacitor of plate separation d is charged to a potential difference V0. A dielectric slab of thickness d and dielectric constant is introduced between the plates while the battery remains connected to the plates. (a) Show that the ratio of energy stored after the dielectric is introduced to the energy stored in the empty capacitor is U/U0 = . (b) Give a physical explanation for this increase in stored energy. (c) What happens to the charge on the capacitor? Note: This situation is not the same as in Example 25.5, in which the battery was removed from the circuit before the dielectric was introduced.