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Chapter 26: Problem 1

(a) How much charge is on each plate of a \(4.00-\mu \mathrm{F}\) capacitor when it is connected to a \(12.0-\mathrm{V}\) battery? (b) If this same capacitor is connected to a \(1.50-\mathrm{V}\) battery, what charge is stored?

Short Answer

Expert verified
The charge on the capacitor when connected to a 12.0-Volt battery is \(4.8 \times 10^{-5} C\). When the same capacitor is connected to a 1.50-Volt battery, the charge is \(6 \times 10^{-6} C\).

Step by step solution

01

Calculating the charge for the 12.0-Volt battery

To find the charge when the capacitor is connected to a 12.0-Volt battery, we substitute the values into the formula \(Q = CV\) . The capacitance (C) is \(4.00碌F = 4.00 \times 10^{-6} F\) (1 碌F = \( 10^{-6} F\) ) and the voltage (V) = 12.0 V. Hence, Q = \(4.00 \times 10^{-6} F \times 12.0 V\).
02

Repeat step one for the 1.50-Volt battery

To find the charge when connected to a 1.50-Volt battery, we apply the formula as we did in step one, only this time substituting 1.50 V for the voltage. Hence, Q = \(4.00 \times 10^{-6} F \times 1.50 V\).

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Most popular questions from this chapter

A small rigid object carries positive and negative \(3.50-\mathrm{nC}\) charges. It is oriented so that the positive charge has coordinates \((-1.20 \mathrm{mm}, 1.10 \mathrm{mm})\) and the negative charge is at the point \((1.40 \mathrm{mm},-1.30 \mathrm{mm}) .\) (a) Find the electric dipole moment of the object. The object is placed in an electric field \(\quad \mathbf{E}=(7800 \mathbf{i}-4900 \hat{\mathbf{j}}) \mathrm{N} / \mathrm{C} .\) (b) Find the torque acting on the object. (c) Find the potential energy of the object-ficld system when the object is in this orientation. (d) If the oricntation of the object can change, find the difference between the maximum and minimum potential energies of the system.

Find the capacitance of the Earth. (Suggestion: The outer conductor of the "spherical capacitor" may be considered as a conducting sphere at infinity where \(V\) approaches zero.)

A wafer of titanium dioxide \((\kappa=173)\) of area \(1.00 \mathrm{cm}^{2}\) has a thickness of \(0.100 \mathrm{mm}\). Aluminum is evaporated on the parallel faces to form a parallel-plate capacitor. (a) Calculate the capacitance. (b) When the capacitor is charged with a \(12.0-\mathrm{V}\) battery, what is the magnitude of charge delivered to each plate? (c) For the situation in part (b), what are the free and induced surface charge densities? (d) What is the magnitude of the electric field?

Consider three capacitors \(C_{1}, C_{2}, C_{3},\) and a battery. If \(C_{1}\) is connected to the battery, the charge on \(C_{1}\) is \(30.8 \mu \mathrm{C}\). Now \(C_{1}\) is disconnected, discharged, and connected in series with \(C_{2} .\) When the series combination of \(C_{2}\) and \(C_{1}\) is connected across the battery, the charge on \(C_{1}\) is \(23.1 \mu \mathrm{C} .\) The circuit is disconnected and the capacitors discharged. Capacitor \(C_{3},\) capacitor \(C_{1},\) and the battery are connected in series, resulting in a charge on \(C_{1}\) of \(25.2 \mu \mathrm{C} .\) If, after being disconnected and discharged, \(C_{1}, C_{2},\) and \(C_{3}\) are connected in series with one another and with the battery, what is the charge on \(C_{1} ?\)

Two conducting spheres with diameters of \(0.400 \mathrm{m}\) and \(1.00 \mathrm{m}\) are separated by a distance that is large compared with the diameters. The spheres are connected by a thin wire and are charged to \(7.00 \mu \mathrm{C}\). (a) How is this total charge shared between the spheres? (Ignore any charge on the wire.) (b) What is the potential of the system of spheres when the reference potential is taken to be \(V=0\) at \(r=\infty ?\)
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