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Chapter 17: Problem 55

A parallel plate capacitor is charged by connecting it to a \(12-V\) battery. The battery is then disconnected from the capacitor. The plates are then pulled apart so the spacing between the plates is increased. What is the effect (a) on the electric field between the plates? (b) on the potential difference between the plates?

Short Answer

Expert verified
Short Answer: The electric field (E2) between the plates after pulling them apart could be larger, smaller, or equal to the initial electric field (E1) depending on the ratio between the final potential difference (V2) and the final distance (d2) between the plates. The potential difference (V2) between the plates after disconnecting the battery and pulling the plates apart will be greater than the initial potential difference (V) due to the decrease in capacitance as the distance increases and the conservation of charge.

Step by step solution

01

Calculating initial capacitance

Since we know that the capacitance formula is C = εA/d, we can denote the initial capacitance as C1 = εA/d1.
02

Calculating initial charge

By using the charge-voltage relationship, we can find the initial charge: Q1 = C1V, where V is the voltage of the battery (12V).
03

Calculating initial electric field

We can find the initial electric field using the formula: E1 = V/d1.
04

Conservation of charge

The charge will be conserved when the capacitor is disconnected from the battery, so Q1 = Q2.
05

Calculating final capacitance

As we pull the plates apart, we change the distance between them, so the final capacitance is given by C2 = εA/d2, where d2 is the new distance.
06

Calculating final potential difference

Using the charge-voltage relationship and the conservation of charge, we can find the final potential difference: Q2 = C2V2, so V2 = Q2/C2.
07

Calculating final electric field

We can now find the final electric field using the formula: E2 = V2/d2. Now we can answer the questions: (a) Since E1 = V/d1 and E2 = V2/d2, and because V2 > V and d2 > d1, it's not possible to give a definitive answer with the information given. The electric field E2 could be larger, smaller, or equal to E1 depending on the ratio between V2 and d2. (b) From the final potential difference expression, V2 = Q2/C2, we know that the capacitance decreases as the distance d2 increases. Since Q2 is constant, this means that the potential difference V2 will be greater than the initial potential difference V.

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Most popular questions from this chapter

Two metal spheres are separated by a distance of \(1.0 \mathrm{cm}\) and a power supply maintains a constant potential difference of \(900 \mathrm{V}\) between them. The spheres are brought closer to one another until a spark flies between them. If the dielectric strength of dry air is $3.0 \times 10^{6} \mathrm{V} / \mathrm{m},$ what is the distance between the spheres at this time?
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