For a parallel plate capacitor 1. Doubling the separation between the plates, doubles the capacitance. 2. Introducing a dielectric of greater permittivity, increases the capacitance. 3. Increasing the area of the plates, increase the capacitance.

For a parallel plate capacitor 1. Doubling the separation between the plates, doubles the capacitance. 2. Introducing a dielectric of greater permittivity, increases the capacitance. 3. Increasing the area of the plates, increase the capacitance.

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter8: Capacitance

Section: Chapter Questions

Problem 82CP: A parallel-plate capacitor is filled with two dielectrics, as shown below. When the plate area is A...

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The plates of an empty parallel-plate capacitor of capacitance 5.0 pF are 2.0 mm apart. What is the area of each plate?
Show that for a given dielectric material the maximum energy a parallel plate capacitor can store is directly proportional to the volume of dielectric (Volume = A. d). Note that the applied voltage is limited by the dielectric strength.
A parallel-plate capacitor is made of two square plates 25 cm on a side and 1.0 mm apart. The capacitor is connected to a 50.0-V battery. With the battery still connected, the plates are pulled apart to a separation of 2.00 mm. What are the energies stored in the capacitor before and after the plates are pulled farther apart? Why does the energy decrease even though work is done in separating 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.
Give the reason why a dielectric material increases capacitance compared with what it would be with air between the plates of a capacitor. What is the independent reason that a dielectric material also allows a greater voltage to be applied to a capacitor? (The dielectric thus increases C and permits a greater V.)
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?
Discuss how the energy stored in an empty but charged capacitor changes when a dielectric is inserted if (a) the capacitor is isolated so that its charge does not change; (b) the capacitor remains connected to a battery so that the potential difference between its plates does not change.
A parallel-plate capacitor is filled with two dielectrics, as shown below. When the plate area is A and separation between plates is d, show that the capacitance is given by C=0Adk1+k22 C=0Adk1+k22 C=0Adk1+k22
You charge a parallel-plate capacitor, remove it from the battery, and prevent the wires connected to the plates from touching each other. When you increase the plate separation, do the following quantities (a) increase, (b) decrease, or (c) stay the same? (i) C (ii) Q (iii) E between the plates (iv) V
What total capacitances can you make by connecting a 5.00 F and an 8.00 F capacitor together?
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?