PHYS 2212 Module 4 Self Assessment Practice Problems

4.1
In one type of computer keyboard, each key holds a small metal plate that acts as one plate of a parallel-plate, air-filled capacitor. When a key is pressed, the plate separation decreases, increasing the capacitance; electronic circuitry detects this change and registers the key press. In one particular keyboard, each metal plate has an area of 48.0 mm², and the plates are separated by 0.740 mm before the key is pressed.
(a) Calculate the capacitance before the key is pressed.
(b) If the circuitry can detect a change in capacitance of 0.290 pF, how far must the key be pressed before the circuitry detects it?
Answer: (a) 0.574 pF (b) 0.25 mm

4.2
A spherical capacitor consists of two concentric spherical shells separated by vacuum, with the outer shell having a radius of 4.60 cm. When connected to a potential difference of 210 V, the capacitor holds a charge of 3.10 nC.
(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

4.3
A spherical capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has a radius of 15.0 cm, and the capacitance of the system is 116 pF.
(a) What is the radius of the outer sphere?
(b) The potential difference between the two spheres is 220 V. What is the magnitude of the charge on each sphere?
Answer: (a) 17.7 cm (b) 25.5 nC

4.4
A parallel-plate, air-filled capacitor with circular plates has a capacitance of 5.00 pF. It will be used in a circuit where it may be subjected to potentials of up to 100 V, and the electric field between its plates must never exceed 1.00 × 10⁴ V/m. As a budding electrical engineer for Live-Wire Electronics, your task is to design this capacitor by determining the radius and separation of its plates.
(a) What must the radius of the plates be?
(b) What must the separation of the plates be?
(c) Find the maximum charge these plates can hold.
Answer: (a) 4.2 cm (b) 1.0 cm (c) 500 pC

4.5
Two capacitors, C₁ = 3.00 µF and C₂ = 5.40 µF, are connected in parallel with each other. The potential difference across the combination, V_ab, is 54.0 V.
(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

4.6
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 V_ab. Once the capacitors are fully charged, the charge on C₂ is 30.0 µC.
(a) What are the charges on capacitors C₁ and C₃?
(b) What is the applied voltage V_ab?
Answer: (a) 60 µC and 90 µC (b) 28 V

4.7
The figure shows a system of four capacitors, where the potential difference between points a and b 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

4.8
For the capacitor network shown in the figure, the potential difference between points a and b 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

4.9
A parallel-plate capacitor has square plates, 8.00 cm on each side, separated by 3.60 mm. The space between the plates is completely filled with two square dielectric slabs, 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). The potential difference between the plates is 78.0 V. How much electrical energy is stored in the capacitor?
Answer: 167 nJ

4.9

A parallel-plate capacitor has square plates, 8.00 cm on each side, separated by 3.60 mm. The space between the plates is completely filled with two square dielectric slabs, 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).

The potential difference between the plates is 78.0 V. How much electrical energy is stored in the capacitor?

Answer: 167 nJ

4.10
A parallel-plate capacitor is made from two plates, 12.0 cm on each side and 4.50 mm apart. As shown in the figure, half of the space between the plates is filled with Plexiglas (dielectric constant 3.40), and the other half contains only air. The plates are connected across an 18.0 V battery.
(a) What is the capacitance of this combination?
(b) How much energy is stored in the capacitor?
(c) Suppose the Plexiglas is removed, with nothing else changed. How much energy will be stored in the capacitor?
Answer: (a) 62.3 pF (b) 10.1 nJ (c) 4.59 nJ