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Chapter 23: Problem 11

Two capacitors contain equal amounts of energy, yet one has twice the capacitance. How do their voltages compare?

Short Answer

Expert verified
The voltage of the capacitor with twice the capacitance is \(\frac{1}{\sqrt{2}}\) times the voltage of the other capacitor.

Step by step solution

01

Determine the relationship between energy, capacitance and voltage

The energy stored in a capacitor can be calculated using the formula \(E = \frac{1}{2} C V^2\). In our case, the energies stored in capacitor A and capacitor B are equal. Thus, we can write the equation as \(\frac{1}{2} C_A V_A^2 = \frac{1}{2} C_B V_B^2\). Simplifying this equation, we get \(C_A V_A^2 = C_B V_B^2\)
02

Substitute the value of the capacitance of capacitor B

We know that the capacitance of capacitor B is twice that of capacitor A, i.e., \(C_B = 2 C_A\). Substituting this in the previous equation, we get \(C_A V_A^2 = 2 C_A V_B^2\). We can divide both sides by \(C_A\) to get \(V_A^2 = 2 V_B^2\)
03

Find the relationship between the voltages

Finally, to find the voltage relation, take the square root of the previous equation to get \(V_A = \sqrt{2} V_B\). This means that the voltage of capacitor A is \(\sqrt{2}\) times the voltage of capacitor B.

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Most popular questions from this chapter

Does the capacitance describe the maximum amount of charge a capacitor can hold, in the same way that a bucket's capacity describes the maximum amount of water it can hold? Explain.

A parallel-plate capacitor with 1.1 -mm plate spacing has \(\pm 2.3 \mu \mathrm{C}\) on its plates when charged to \(150 \mathrm{V}\). What's the plate area?

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