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Chapter 24: Problem 23

A parallel plate capacitor is constructed from two plates of different areas. If this capacitor is initially uncharged and then connected to a battery, how will the amount of charge on the big plate compare to the amount of charge on the small plate?

Short Answer

Expert verified
Answer: The charges on the big and small plates of a parallel plate capacitor will be equal and opposite, irrespective of the sizes of the plates, because of the homogenous electric field between them and charge conservation in a closed circuit.

Step by step solution

01

Understanding the Capacitor

A parallel plate capacitor consists of two conducting plates separated by a distance. When connected to a battery, one plate gains positive charge, and the other plate gains an equal amount of negative charge. The charges on both plates will be equal and opposite.
02

Charging Process

When the parallel plate capacitor is connected to a battery, the battery provides positive charges to one plate and takes away an equal amount of charges from the other plate, making it negatively charged. This process continues until the potential difference across the plates becomes equal to the battery's EMF.
03

Amount of Charge on Plates

In the given exercise, the capacitor has two plates with different areas. However, charges will still distribute uniformly on both plates as the electric field between them is homogenous. Therefore, the charges on both plates will be equal and opposite, keeping in mind that the total charges on the connected components in a closed circuit remain conserved.
04

Conclusion

The charges on the big and small plates of a parallel plate capacitor will be equal and opposite, irrespective of the sizes of the plates, because of the homogenous electric field between them and charge conservation in a closed circuit.

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Most popular questions from this chapter

A parallel plate capacitor of capacitance \(C\) has plates of area \(A\) with distance \(d\) between them. When the capacitor is connected to a battery of potential difference \(V\), it has a charge of magnitude \(Q\) on its plates. While the capacitor is connected to the battery, the distance between the plates is decreased by a factor of \(3 .\) The magnitude of the charge on the plates and the capacitance are then a) \(\frac{1}{3} Q\) and \(\frac{1}{3} C\). c) \(3 Q\) and \(3 C\). b) \(\frac{1}{3} Q\) and \(3 C\). d) \(3 Q\) and \(\frac{1}{3} C\).

Design a parallel plate capacitor with a capacitance of \(47.0 \mathrm{pF}\) and a capacity of \(7.50 \mathrm{nC}\). You have available conducting plates, which can be cut to any size, and Plexiglas sheets, which can be cut to any size and machined to any thickness. Plexiglas has a dielectric constant of 3.40 and a dielectric strength of \(4.00 \cdot 10^{7} \mathrm{~V} / \mathrm{m}\). You must make your capacitor as compact as possible. Specify all relevant dimensions. Ignore any fringe field at the edges of the capacitor plates.

The capacitance of a spherical capacitor consisting of two concentric conducting spheres with radii \(r_{1}\) and \(r_{2}\) \(\left(r_{2}>r_{1}\right)\) is given by \(C=4 \pi \epsilon_{0} r_{1} r_{2} /\left(r_{2}-r_{1}\right) .\) Suppose that the space between the spheres, from \(r,\) up to a radius \(R\) \(\left(r_{1}

Does it take more work to separate the plates of a charged parallel plate capacitor while it remains connected to the charging battery or after it has been disconnected from the charging battery?

Fifty parallel plate capacitors are connected in series. The distance between the plates is \(d\) for the first capacitor, \(2 d\) for the second capacitor, \(3 d\) for the third capacitor, and so on. The area of the plates is the same for all the capacitors. Express the equivalent capacitance of the whole set in terms of \(C_{1}\) (the capacitance of the first capacitor).
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