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Chapter 17: Problem 80

The capacitor of Problem 79 is initially charged to a \(150-\mathrm{V}\) potential difference. The plates are then physically separated by another \(0.750 \mathrm{mm}\) in such a way that none of the charge can leak off the plates. Find (a) the new capacitance and (b) the new energy stored in the capacitor. Explain the result using conservation of energy.

Short Answer

Expert verified
(b) What is the new energy stored in the capacitor after the additional separation?

Step by step solution

01

Recall the capacitance formula for a parallel plate capacitor

The capacitance of a parallel plate capacitor can be calculated using the following formula: $$C=\frac{\epsilon A}{d}$$ where \(C\) is the capacitance, \(\epsilon\) is the vacuum permittivity constant, \(A\) is the area of one of the capacitor plates, and \(d\) is the distance between the plates.
02

Calculate the initial capacitance

Initially, the capacitor is charged to a potential difference of 150 V, and we know its capacitance from Problem 79 (which we denote as \(C_{1}\)). Given the potential difference \(V_{1}\) and the charge \(Q\) (which remains constant when the plates are separated), we can calculate the initial energy stored in the capacitor using the formula: $$U_{1}=\frac{1}{2}C_{1}V_{1}^2$$
03

Calculate the new distance between the plates

The plates of the capacitor are separated by an additional 0.750 mm, so we need to calculate the new distance between them, denoted as \(d_{2}\). Let \(d_{1}\) be the initial distance between the plates. We have: $$d_{2}=d_{1}+0.750\times10^{-3}$$
04

Calculate the new capacitance

To find the new capacitance, \(C_{2}\), we use the formula for the capacitance of a parallel plate capacitor, with the new distance \(d_2\) calculated in the previous step: $$C_{2}=\frac{\epsilon A}{d_{2}}$$
05

Calculate the new potential difference

We know that the charge on the capacitor remains constant. Using the relationship between charge, capacitance, and potential difference, \(Q=C_1V_1=C_2V_2\), we can calculate the new potential difference, \(V_2\): $$V_{2}=\frac{C_{1}}{C_{2}}V_{1}$$
06

Calculate the new energy stored in the capacitor

We can now find the new energy stored in the capacitor, \(U_2\), using the new capacitance, \(C_2\), and the new potential difference, \(V_2\). The formula for the energy stored in a capacitor is: $$U_{2}=\frac{1}{2}C_{2}V_{2}^2$$
07

Explain the result using conservation of energy

According to the conservation of energy principle, the initial energy stored in the capacitor should be equal to the new energy stored plus the work done to separate the plates. Mathematically, this can be written as: $$U_{1}=U_{2}+W$$ Where \(W\) is the work done to separate the plates. This can be used to analyze the results obtained in the previous steps and to verify the conservation of energy in this case.

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Most popular questions from this chapter

A parallel plate capacitor is connected to a \(12-\mathrm{V}\) battery. While the battery remains connected, the plates are pushed together so the spacing is decreased. What is the effect on (a) the potential difference between the plates? (b) the electric field between the plates? (c) the magnitude of charge on the plates?
A point charge \(q=-2.5 \mathrm{nC}\) is initially at rest adjacent to the negative plate of a capacitor. The charge per unit area on the plates is $4.0 \mu \mathrm{C} / \mathrm{m}^{2}\( and the space between the plates is \)6.0 \mathrm{mm} .$ (a) What is the potential difference between the plates? (b) What is the kinetic energy of the point charge just before it hits the positive plate, assuming no other forces act on it?
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