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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 渭C.

Step by step solution

01

Convert capacitance to farads

First, we need to convert the given capacitance from microfarads (渭F) to farads (F). To do this, we use the conversion: 1 渭F = 10^{-6} F So, 2.0 渭F = 2.0 脳 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: \(Q = C V\) \(Q = (2.0 脳 10^{-6} \text{ F})(9.0 \text{ V})\)
03

Calculate the charge on the capacitor

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

Write the final answer

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

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