PHYS 2212 Module 4 Self Assessment Practice Problems

Module 4 Self Assessment Practice Problems

In one type of computer keyboard, each key holds a small metal plate that serves as one plate of a parallel-plate, air-filled capacitor. When the key is depressed, the plate separation decreases and the capacitance increases. Electronic circuitry detects the change in capacitance and thus detects that the key has been pressed. In one particular keyboard, the area of each metal plate is 48.0 mm2, and the separation between the plates is 0.740 mm before the key is depressed.
(a) Calculate the capacitance before the key is depressed.
(b) If the circuitry can detect a change in capacitance of 0.290 pF, how far must the key be depressed before the circuitry detects its depression?
Answer:   (a) 0.574 pF  (b) 0.49 mm
A spherical capacitor contains a charge of 3.10 nC when connected to a potential difference of 210 V. Its plates are separated by vacuum and the radius of the outer shell is 4.60 cm.
(a) Calculate the capacitance.
(b) Calculate the radius of the inner sphere.
(c) Calculate the electric field just outside the surface of the inner sphere.
Answer:   (a) 14.8 pF  (b) 3.4 cm  (c) 2.42 x 104 V/m
A spherical capacitor is formed from two concentric, spherical, conducting shells separated by vacuum. The inner sphere has radius 15.0 cm and the capacitance is 116 pF.
(a) What is the radius of the outer sphere?
(b) If the potential difference between the two spheres is 220 V, what is the magnitude of charge on each sphere?
Answer:  (a) 17.7 cm  (b) 25.5 nC
A 5.00-pF parallel-plate air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to potentials of up to 1.00 x 102 V. The electric field between the plates is to be no greater than 1.00 x 104 V/m. As a budding electrical engineer for Live-Wire Electronics, your task is to design the capacitor by finding what its physical dimensions and separation must be.
(a) What must the radius of the plates in the designed capacitor be?
(b) What must the separation of the plates in the designed capacitor be?
(c) Find the maximum charge these plates can hold.
Answer:   (a) 4.2 cm  (b) 9.8 mm  (c) 500 pC
Two capacitors are connected parallel to each other. Let C1 = 3.00 µF, C2 = 5.40 µF be their capacitances, and Vab = 54.0 V the potential difference across the system.
(a) Calculate the potential difference across each capacitor.
(b) Calculate the charge on each capacitor.
Answer:  (a) 54 V and 54 V   (b) 162 µC and 292 µC

In the figure, C1 = 6.00 µF, C2 = 3.00 µF, and C3 = 5.00 µF. The capacitor network is connected to an applied potential Vab. After the charges on the capacitors have reached their final values, the charge on C2 is 30.0 µC.
(a) What are the charges on capacitors C1 and C3?
(b) What is the applied voltage Vab?
Answer:  (a) 60 µC and 90 µC  (b) 28 V

The figure shows a system of four capacitors, where the potential difference across ab is 50.0 V.
(a) Find the equivalent capacitance of this system between a and b.
(b) How much charge is stored by this combination of capacitors?
(c) How much charge is stored in the 10.0-µF capacitor?
(d) How much charge is stored in the 9.0-µF capacitor?
Answer:   (a) 3.47 µF  (b) 174 µC  (c) 174 µC  (d) 174 µC

For the capacitor network shown in the figure, the potential difference across ab is 12.0 V.
(a) Find the total energy stored in this network.
(b) Find the energy stored in the 4.80-µF capacitor.
Answer:   (a) 1.58 x 10-4 J   (b) 72 µJ
A parallel-plate capacitor has square plates that are 8.00 cm on each side and 3.60 mm apart. The space between the plates is completely filled with two square slabs of dielectric, each 8.00 cm on a side and 1.80 mm thick. One slab is pyrex glass (k = 5.49) and the other is polystyrene (k = 2.56). If the potential difference between the plates is 78.0 V, how much electrical energy is stored in the capacitor?
Answer:  167 nJ

A parallel-plate capacitor is made from two plates 12.0 cm on each side and 4.50 mm apart. Half of the space between these plates contains only air, but the other half is filled with Plexiglas of dielectric constant 3.40. An 18.0 V battery is connected across the plates.
(a) What is the capacitance of this combination?
(b) How much energy is stored in the capacitor?
(c) If we remove the Plexiglas, but change nothing else, how much energy will be stored in the capacitor?
Answer:   (a) 62.3 pF  (b) 10.1 nJ  (c) 4.59 nJ