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Chapter 17: Problem 58

A parallel plate capacitor has a capacitance of \(1.20 \mathrm{nF}\) and is connected to a \(12-\mathrm{V}\) battery. (a) What is the magnitude of the charge on each plate? (b) If the plate separation is doubled while the plates remain connected to the battery, what happens to the charge on each plate and the electric field between the plates?

Short Answer

Expert verified
Answer: When the plate separation is doubled while the plates remain connected to the battery, the charge on each plate is halved, and the electric field between the plates is also halved. The magnitude of the charge on each plate initially is half of (1.20 nF) × (12 V).

Step by step solution

01

Calculate the initial charge on each plate

Begin by calculating the charge on each plate using the capacitance formula: \(C = \frac{Q}{V}\). Remember, we have a capacitance of \(C = 1.20 \mathrm{nF}\) and a connected battery with a voltage of \(V = 12 \mathrm{V}\). Rearrange the equation to find the charge, \(Q\): \(Q = C \times V\) Now, plug in the given values: \(Q = (1.20 \mathrm{nF}) \times (12 \mathrm{V})\)
02

Calculate the charge on each plate

Next, calculate the charge on each plate. Since the total charge is evenly distributed between the two plates, the charge on each plate is half of the total charge: \(Q_\text{each} = \frac{Q}{2}\)
03

Determine the effect of doubling the plate separation on the charge

The plates remaining connected to the battery means that the voltage across the capacitor remains constant. Because the capacitance formula is \(C = \frac{Q}{V}\), and \(V\) remains constant, if the capacitance changes, the charge (\(Q\)) must also change accordingly. Doubling the plate separation has an effect on the capacitance. The capacitance is inversely proportional to the distance between the plates: \(C \propto \frac{1}{d}\) If the distance is doubled, the capacitance is halved. Since the capacitance is half of its original value, the charge on each plate will also be halved.
04

Determine the effect of doubling the plate separation on the electric field

To determine the effect of doubling the plate separation on the electric field, we can use the formula: \(E = \frac{V}{d}\) If we double the separation, \(d\), we can see that the electric field will be half of its original value. Therefore, the electric field between the plates halves when the plate separation is doubled while the plates remain connected to the battery. To summarize: (a) The magnitude of the charge on each plate is half of \(Q = (1.20 \mathrm{nF}) \times (12 \mathrm{V})\) (b) If the plate separation is doubled while the plates remain connected to the battery, the charge on each plate is halved, and the electric field between the plates is also halved.

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Most popular questions from this chapter

An array of four charges is arranged along the \(x\) -axis at intervals of 1.0 \(\mathrm{m}\). (a) If two of the charges are \(+1.0 \mu \mathrm{C}\) and two are \(-1.0 \mu \mathrm{C},\) draw a configuration of these charges that minimizes the potential at \(x=0 .\) (b) If three of the charges are the same, $q=+1.0 \mu \mathrm{C},\( and the charge at the far right is \)-1.0 \mu \mathrm{C},$ what is the potential at the origin?
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