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Question

Two 2.0-cm-diameter disks spaced 2.0 mm apart form a parallel-plate capacitor. The electric field between the disks is 5.0×105V/m5.0 \times 10^{5} \mathrm{V} / \mathrm{m}. a. What is the voltage across the capacitor? b. How much charge is on each disk? c. An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.0×107m/s2.0 \times 10^{7} \mathrm{m} / \mathrm{s}. What was the electron's speed as it left the negative plate?

Solution

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Answered 2 years ago
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Approach:

We are given the following data:

D=2 cmd=2 mmE=5105 V mvf=2107 m s\begin{aligned} &D=2\text{ cm}\\ &d=2\text{ mm}\\ &E=5\cdot 10^5\dfrac{\text{ V}}{\text{ m}}\\ &v_f=2\cdot 10^7\dfrac{\text{ m}}{\text{ s}} \end{aligned}

To get the requested data, let's look at how they are related to the given data.

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