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Chapter 17: Problem 52

The plates of a 15.0 - \(\mu \mathrm{F}\) capacitor have net charges of $+0.75 \mu \mathrm{C}\( and \)-0.75 \mu \mathrm{C},$ respectively. (a) What is the potential difference between the plates? (b) Which plate is at the higher potential?

Short Answer

Expert verified
Answer: The potential difference between the plates is 0.05 V, and the plate with the higher potential is the one with the +0.75 μC charge.

Step by step solution

01

Find the potential difference between the plates

Using the given information, we can use the formula for the capacitance of a capacitor, which is \(C = \frac{Q}{V}\), where \(C\) is the capacitance, \(Q\) is the charge, and \(V\) is the potential difference. We can rearrange this formula to solve for the potential difference: \(V = \frac{Q}{C}\). Insert the given values: \(C = 15.0 \, \mu \mathrm{F} = 15.0 \times 10^{-6} \mathrm{F}\) \(Q = 0.75 \, \mu \mathrm{C} = 0.75 \times 10^{-6} \mathrm{C}\) Calculate the potential difference: \(V = \frac{0.75 \times 10^{-6} \mathrm{C}}{15.0 \times 10^{-6} \mathrm{F}}\)
02

Calculate the potential difference

Now we can calculate the potential difference: \(V = \frac{0.75 \times 10^{-6} \mathrm{C}}{15.0 \times 10^{-6} \mathrm{F}} = 0.05 \mathrm{V}\) So the potential difference between the plates is 0.05 V.
03

Determine the plate with the higher potential

From the problem statement, we have the charges on the plates as +0.75 \(\mu\)C and -0.75 \(\mu\)C. The plate with the higher potential will be the plate with the positive charge, which is the plate with a charge of +0.75 \(\mu\)C. To summarize: a) The potential difference between the plates is 0.05 V. b) The plate with the higher potential is the one with the +0.75 \(\mu\)C charge.

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Most popular questions from this chapter

A parallel plate capacitor has a charge of \(0.020 \mu \mathrm{C}\) on each plate with a potential difference of \(240 \mathrm{V}\) The parallel plates are separated by \(0.40 \mathrm{mm}\) of bakelite. What is the capacitance of this capacitor?
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?
(a) Calculate the capacitance per unit length of an axon of radius $5.0 \mu \mathrm{m} \text { (see Fig. } 17.14) .$ The membrane acts as an insulator between the conducting fluids inside and outside the neuron. The membrane is 6.0 nm thick and has a dielectric constant of \(7.0 .\) (Note: The membrane is thin compared with the radius of the axon, so the axon can be treated as a parallel plate capacitor.) (b) In its resting state (no signal being transmitted), the potential of the fluid inside is about \(85 \mathrm{mV}\) lower than the outside. Therefore, there must be small net charges \(\pm Q\) on either side of the membrane. Which side has positive charge? What is the magnitude of the charge density on the surfaces of the membrane?
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.
Suppose a uniform electric field of magnitude \(100.0 \mathrm{N} / \mathrm{C}\) exists in a region of space. How far apart are a pair of equipotential surfaces whose potentials differ by \(1.0 \mathrm{V} ?\)
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