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Chapter 17: Problem 51

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?

Short Answer

Expert verified
Answer: The magnitude of the charge on each plate of the capacitor is 18.0 μC.

Step by step solution

01

Convert capacitance to farads

First, we need to convert the given capacitance from microfarads (μF) to farads (F). To do this, we use the conversion: 1 μF = 10^{-6} F So, 2.0 μF = 2.0 × 10^{-6} F
02

Plug the values into the formula

Now, we have all the required values in their respective units. Let's plug them into the formula for the charge on the capacitor: \(Q = C V\) \(Q = (2.0 × 10^{-6} \text{ F})(9.0 \text{ V})\)
03

Calculate the charge on the capacitor

Now, let's calculate the charge on the capacitor: \(Q = (2.0 × 10^{-6} \text{ F})(9.0 \text{ V}) = 18.0 × 10^{-6} \text{ C}\)
04

Write the final answer

The magnitude of the charge on each plate of the capacitor is: \(Q = 18.0 × 10^{-6} \text{ C}\) Or, simply: \(Q = 18.0 \text{ μC}\)

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Most popular questions from this chapter

To make a parallel plate capacitor, you have available two flat plates of aluminum (area \(120 \mathrm{cm}^{2}\) ), a sheet of paper (thickness $=0.10 \mathrm{mm}, \kappa=3.5),\( a sheet of glass (thickness \)=2.0 \mathrm{mm}, \kappa=7.0),\( and a slab of paraffin (thickness \)=10.0 \mathrm{mm}, \kappa=2.0) .$ (a) What is the largest capacitance possible using one of these dielectrics? (b) What is the smallest?
The potential difference across a cell membrane is \(-90 \mathrm{mV} .\) If the membrane's thickness is \(10 \mathrm{nm},\) what is the magnitude of the electric field in the membrane? Assume the field is uniform.
A cell membrane has a surface area of \(1.0 \times 10^{-7} \mathrm{m}^{2},\) a dielectric constant of \(5.2,\) and a thickness of \(7.5 \mathrm{nm}\) The membrane acts like the dielectric in a parallel plate capacitor; a layer of positive ions on the outer surface and a layer of negative ions on the inner surface act as the capacitor plates. The potential difference between the "plates" is \(90.0 \mathrm{mV}\). (a) How much energy is stored in this capacitor? (b) How many positive ions are there on the outside of the membrane? Assume that all the ions are singly charged (charge +e).
A parallel plate capacitor is connected to a battery. The space between the plates is filled with air. The electric field strength between the plates is \(20.0 \mathrm{V} / \mathrm{m} .\) Then, with the battery still connected, a slab of dielectric \((\kappa=4.0)\) is inserted between the plates. The thickness of the dielectric is half the distance between the plates. Find the electric field inside the dielectric.
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 air. (a) What is the capacitance for this capacitor? (b) What is the area of a single plate? (c) At what voltage will the air between the plates become ionized? Assume a dielectric strength of \(3.0 \mathrm{kV} / \mathrm{mm}\) for air.
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