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Chapter 24: Problem 9

A parallel plate capacitor is connected to a battery for charging. After some time, while the battery is still connected to the capacitor, the distance between the capacitor plates is doubled. Which of the following is (are) true? a) The electric field between the plates is halved. b) The potential difference of the battery is halved. c) The capacitance doubles. d) The potential difference across the plates does not change. e) The charge on the plates does not change.

Short Answer

Expert verified
(Select all that apply.) a) The electric field between the plates is halved. b) The potential difference of the battery is halved. c) The capacitance doubles. d) The potential difference across the plates does not change. e) The charge on the plates does not change. Answer: a) The electric field between the plates is halved, d) The potential difference across the plates does not change, and e) The charge on the plates does not change are true statements.

Step by step solution

01

a) The electric field between the plates is halved

After doubling the distance, we can use the electric field formula to find out how the electric field changes: E_new = V/(2d), which is equal to (1/2) * V/d. This shows that the electric field (E_new) is indeed halved as compared to the initial electric field (E).
02

b) The potential difference of the battery is halved

The potential difference across the battery remains constant as it is not affected by the change in distance between the plates. Therefore, this statement is false.
03

c) The capacitance doubles

Using the capacitance formula for a parallel plate capacitor, we can see how the capacitance changes when the distance between the plates is doubled: C_new = ε₀ * A/(2d). This is equal to (1/2) * ε₀ * A/d which is half of the initial capacitance, not double. So, this statement is false.
04

d) The potential difference across the plates does not change

As the capacitor is still connected to the battery, the potential difference across the plates remains the same as it is equal to the potential difference of the battery. Therefore, this statement is true.
05

e) The charge on the plates does not change

As the potential difference remains the same and the charge formula is Q = CV, since we found out that C is halved, and V remains constant, the charge on the plates will also not change. Therefore, this statement is true. In conclusion, statements a), d), and e) are true, while statements b) and c) are false.

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Most popular questions from this chapter

Considering the dielectric strength of air, what is the maximum amount of charge that can be stored on the plates of a capacitor that are a distance of \(15 \mathrm{~mm}\) apart and have an area of \(25 \mathrm{~cm}^{2}\) ?

Which of the following capacitors has the largest charge? a) a parallel plate capacitor with an area of \(10 \mathrm{~cm}^{2}\) and a plate separation of \(2 \mathrm{~mm}\) connected to a \(10-\mathrm{V}\) battery b) a parallel plate capacitor with an area of \(5 \mathrm{~cm}^{2}\) and a plate separation of \(1 \mathrm{~mm}\) connected to a \(10-\mathrm{V}\) battery c) a parallel plate capacitor with an area of \(10 \mathrm{~cm}^{2}\) and a plate separation of \(4 \mathrm{~mm}\) connected to a \(5-\mathrm{V}\) battery d) a parallel plate capacitor with an area of \(20 \mathrm{~cm}^{2}\) and a plate separation of \(2 \mathrm{~mm}\) connected to a \(20-\mathrm{V}\) battery e) All of the capacitors have the same charge.

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).

Two parallel plate capacitors, \(C_{1}\) and \(C_{2},\) are connected in series to a \(96.0-\mathrm{V}\) battery. Both capacitors have plates with an area of \(1.00 \mathrm{~cm}^{2}\) and a separation of \(0.100 \mathrm{~mm} ;\) \(C_{1}\) has air between its plates, and \(C_{2}\) has that space filled with porcelain (dielectric constant of 7.0 and dielectric strength of \(5.70 \mathrm{kV} / \mathrm{mm}\) ). a) After charging, what are the charges on each capacitor? b) What is the total energy stored in the two capacitors? c) What is the electric field between the plates of \(C_{2} ?\)

The Earth is held together by its own gravity. But it is also a charge-bearing conductor. a) The Earth can be regarded as a conducting sphere of radius \(6371 \mathrm{~km},\) with electric field \(\vec{E}=(-150 . \mathrm{V} / \mathrm{m}) \hat{r}\) at its surface, where \(\hat{r}\) is a unit vector directed radially outward. Calculate the total electrostatic potential energy associated with the Earth's electric charge and field. b) The Earth has gravitational potential energy, akin to the electrostatic potential energy. Calculate this energy, treating the Earth as a uniform solid sphere. (Hint: \(d U=-(G m / r) d m\). c) Use the results of parts (a) and (b) to address this question: To what extent do electrostatic forces affect the structure of the Earth?
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