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Chapter 17: Problem 57

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?

Short Answer

Expert verified
Answer: (a) The potential difference remains constant at 12V, (b) the electric field doubles, and (c) the charge on the plates doubles.

Step by step solution

01

Calculate the initial capacitance

Initially, let's call the capacitance of the capacitor \(C_{1}\). We don't have exact values for \(A\), \(d\), nor the dielectric constant \(\kappa\), but we know that \(C_{1} = \frac{\kappa\varepsilon_{0} A}{d}\).
02

Calculate the new capacitance

Let's assume that the distance between the plates is reduced to half. So, the new distance between the plates becomes \(d_{2}= \frac{1}{2} d\). Since area \(A\) and dielectric constant \(\kappa\) remain the same, the new capacitance \(C_{2}\) can be calculated using the formula \(C_{2}=\frac{\kappa\varepsilon_{0}A}{d_{2}}= \frac{\kappa\varepsilon_{0}A}{\frac{1}{2}d} = 2 \frac{\kappa\varepsilon_{0}A}{d}=2C_{1}\).
03

Calculate the new potential difference

While the battery remains connected, the potential difference across the capacitor remains constant. Therefore, the potential difference between the plates is unaffected and remains \(12\ \text{V}\). #b) Effect on the electric field#
04

Calculate the initial electric field

Using the initial potential difference and the initial distance between the plates, the initial electric field \(E_{1}\) can be calculated using the formula \(E_{1} = \frac{V}{d}\).
05

Calculate the new electric field

Since the potential difference remains constant and the new distance between the plates has been reduced to half, the new electric field \(E_{2}\) can be calculated using the formula \(E_{2}= \frac{V}{d_{2}}= \frac{V}{\frac{1}{2}d} = 2 \frac{V}{d} = 2E_{1}\). Thus, the electric field doubles when the distance is decreased to half. #c) Effect on the magnitude of charge on the plates#
06

Calculate the initial charge

We know that \(Q = CV\), so the initial charge on the plates, \(Q_{1}\), can be computed as \( Q_{1} = C_{1}V\).
07

Calculate the new charge

Since we found that the new capacitance is \(C_{2}=2C_{1}\) and the potential difference remains the same, the new charge on the plates, \(Q_{2}\), can be computed as \(Q_{2}=C_{2}V\). But, \(C_{2}V = 2C_{1}V =2Q_{1}\). Hence, the magnitude of the charge on the plates doubles when the distance is decreased to half.

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Most popular questions from this chapter

A 2.0 - \(\mu\) F capacitor is connected to a \(9.0-\mathrm{V}\) battery. What is the magnitude of the charge on each plate?
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A 4.00 - \(\mu \mathrm{F}\) air gap capacitor is connected to a \(100.0-\mathrm{V}\) battery until the capacitor is fully charged. The battery is removed and then a dielectric of dielectric constant 6.0 is inserted between the plates without allowing any charge to leak off the plates. (a) Find the energy stored in the capacitor before and after the dielectric is inserted. [Hint: First find the new capacitance and potential difference.] (b) Does an external agent have to do positive work to insert the dielectric or to remove the dielectric? Explain.
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