Chapter 30 Homework Solutions
Solutions for the following Self Tutoring and Skill Builder problems are available through Mastering Physics: Energy in Capacitors and Electric Fields, Potential of a Charged Disk, and Potential of Two Charged Spheres.
To access these solutions, click on the View Solution link on each problem view window.30.4.Model: The electric field is the negative of the derivative of the potential function.Solve: From Equation 30.8, the component of the electric field in the s-direction issEdVds.
For the given potential, 2V100 V200 200 VmmxdVdxxExdxdx At x
0 m, Ex
0 V/m , and at x
1 m, E
200 (1) V/m
200 V/m.Assess: The potential increases with x, so the electric field must point in the x direction.
30.32.Solve: The energy density is .u 21E02uE
(b) Equation 30.3 gives the potential difference between two points in space:
Assess: E is positive for negative x but negative for positive x, so the potential defference depends on the square of the positions.
30.47.Model: Conductors connected by a conducting wire are at the same potential.Visualize:
Charge flows from sphere 1 to sphere 2, when the wire is connected.
Once connected, the entire system is one large conductor in electrostatic equilibrium.Consequently, the entire system must be an equipotential
That is, the two spheres must have equal potentials V1
V2 after being connected by the wire.
Solve: The potential of a sphere is 04VQR, so the final charges Q1f and Q2f on the two spheres must be such that 1f2f2f0102014442QQQRRR11f2f2QQ where we used the fact that R2
2R1.It is also true that charge must be conserved, so nC Using Q2f
2Q1f, we find nC from which we finally get Q1f
2 nC and Q2f
30.68.Model: Assume the battery is ideal.
The circuit in Figure P30.68 has been redrawn to show that the six capacitors are arranged in three parallel combinations, each combination being a series combination of two capacitors.Solve: (a) The equivalent capacitance of the two capacitors in series is 12C.The equivalent capacitance of the six capacitors is 32C.
(b) As points a and b are midpoints of identical capacitors, Va
6.0 V.Therefore, the potential difference between points a and b is zero.
Solve: (a) The capacitance of the parallel-plate capacitor is 12221101310.1 m
0.1 m8.8510 C/N m8.8510 F1.010 mACd The electric potential energy stored in the capacitor is 292711111010 C115.6510 J228.8510FQUC (b) There is no change in the charge.The energy change is due to the change in the capacitance.The new capacitance is 00122122AACCdd The amount of energy stored is U.7211.310 J(c) Work was done on the capacitor by the agent pulling the plates apart, thereby adding energy into the system..