Skip to main content
Ch 24: Capacitance and Dielectrics
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 24, Problem 11

A spherical capacitor contains a charge of 3.303.30 nC when connected to a potential difference of 220220 V. If its plates are separated by vacuum and the inner radius of the outer shell is 4.004.00 cm, calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

Verified step by step guidance
1
To find the capacitance of a spherical capacitor, use the formula for capacitance: C = QV, where Q is the charge and V is the potential difference. Substitute the given values: Q = 3.30 imes 10^{-9} \(\text{ C}\) and V = 220 \(\text{ V}\).
For part (b), use the formula for the capacitance of a spherical capacitor: C = 4\(\text{π}\)\(\text{ε}\)_0 rac{r_1 r_2}{r_2 - r_1}, where r_1 is the radius of the inner sphere and r_2 is the radius of the outer shell. Rearrange the formula to solve for r_1.
Substitute the known values into the rearranged formula: C from part (a), r_2 = 0.04 \(\text{ m}\), and ε_0 = 8.85 imes 10^{-12} \(\text{ F/m}\).
For part (c), calculate the electric field just outside the surface of the inner sphere using the formula: E = rac{Q}{4\(\text{π}\)\(\text{ε}\)_0 r_1^2}. Substitute the values for Q, ε_0, and r_1 from part (b).
Ensure all units are consistent, converting cm to m where necessary, and verify each calculation step to ensure accuracy in determining the capacitance, radius, and electric field.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
5m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Capacitance of a Spherical Capacitor

Capacitance is a measure of a capacitor's ability to store charge per unit voltage. For a spherical capacitor, it depends on the radii of the inner and outer spheres and the permittivity of the medium between them. The formula for capacitance in a vacuum is C = 4πε₀(r₁r₂)/(r₂ - r₁), where r₁ and r₂ are the radii of the inner and outer spheres, respectively.
Recommended video:
Guided course
03:11
Capacitance of Spherical Capacitor

Electric Field in a Spherical Capacitor

The electric field in a spherical capacitor is determined by the charge on the inner sphere and the distance from the center. Just outside the surface of the inner sphere, the electric field E can be calculated using E = Q/(4πε₀r₁²), where Q is the charge and r₁ is the radius of the inner sphere. This field is radially outward and decreases with distance.
Recommended video:
Guided course
06:13
Intro to Capacitors

Potential Difference and Charge Relationship

The relationship between potential difference (voltage) and charge in a capacitor is given by Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. This equation helps in determining the capacitance when the charge and voltage are known, and is fundamental in analyzing capacitors in circuits.
Recommended video:
Guided course
05:45
Potential Difference Between Two Charges
Related Practice
Textbook Question

For the system of capacitors shown in Fig. E24.1624.16, find the equivalent capacitance between bb and cc.

2719
views
1
rank
Textbook Question

A 5.005.00-μ\(\mu\)F parallel-plate capacitor is connected to a 12.012.0 V battery. After the capacitor is fully charged, the battery is disconnected without loss of any of the charge on the plates.

(a) A voltmeter is connected across the two plates without discharging them. What does it read?

(b) What would the voltmeter read if (i) the plate separation were doubled; (ii) the radius of each plate were doubled but their separation was unchanged?

2866
views
1
rank
Textbook Question

A parallel-plate air capacitor is to store charge of magnitude 240.0240.0 pC on each plate when the potential difference between the plates is 42.042.0 V.

(a) If the area of each plate is 6.806.80 cm2, what is the separation between the plates?

(b) If the separation between the two plates is double the value calculated in part (a), what potential difference is required for the capacitor to store charge of magnitude 240.0240.0 pC on each plate?

2539
views
Textbook Question

Figure E24.1424.14 shows a system of four capacitors, where the potential difference across ab is 50.050.0 V. How much charge is stored in each of the 10.010.0-μ\(\mu\)F and the 9.09.0-μ\(\mu\)F capacitors?

3272
views
Textbook Question

A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and the outer is positively charged; the magnitude of the charge on each is 10.010.0 pC. The inner cylinder has radius 0.500.50 mm, the outer one has radius 5.005.00 mm, and the length of each cylinder is 18.018.0 cm.

(a) What is the capacitance?

(b) What applied potential difference is necessary to produce these charges on the cylinders?

1975
views
Textbook Question

Figure E24.1424.14 shows a system of four capacitors, where the potential difference across ab is 50.050.0 V. How much charge is stored by this combination of capacitors?

1807
views