1. A pair of oppositely charged parallel plates is separated by 5.51 mm. A potential difference of 614 V exists between the plates. What is the strength of the electric field between the plates? The fundamental charge is 1.602 × 10?19 . Answer in units of V/m.
2. What is the magnitude of the force on an electron between the plates? Answer in units of N.
3. How much work must be done on the electron to move it to the negative plate if it is initially positioned 2.7 mm from the positive plate? Answer in units of J.

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aristocles aristocles · Answer 1 · 2019-11-25T01:54:58+01:00

Answer:

Part a)

[tex]E = 1.11 \times 10^5 N/C[/tex]

Part 2)

[tex]F = 1.78 \times 10^{-14} N[/tex]

Part 3)

[tex]W = 5 \times 10^{-17} J[/tex]

Explanation:

Part 1)

As we know that electric field and potential difference related to each other as

[tex]E = \frac{\Delta V}{x}[/tex]

so we will have

[tex]\Delta V = 614 V[/tex]

[tex]x = 5.51 mm[/tex]

so we have

[tex]E = \frac{614}{5.51 \times 10^{-3}}[/tex]

[tex]E = 1.11 \times 10^5 N/C[/tex]

Part 2)

Charge of an electron

[tex]e = 1.6 \times 10^{-19} C[/tex]

now force is given as

[tex]F = qE[/tex]

[tex]F = (1.6 \times 10^{-19})(1.11 \times 10^5)[/tex]

[tex]F = 1.78 \times 10^{-14} N[/tex]

Part 3)

Work done to move the electron

[tex]W = F.d[/tex]

[tex]W = (1.78 \times 10^{-14})(5.51 - 2.7) \times 10^{-3}[/tex]

[tex]W = 5 \times 10^{-17} J[/tex]