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Oppositely charged parallel plates are separated by 4.67 mm. A potential difference of 600 V exists between the plates. (a) What is the magnitude of the electric field between the plates? ________ N/C (b) What is the magnitude of the force on an electron between the plates? ________ N (c) How much work must be done on the electron to move it to the negative plate if it is initially positioned 2.00 mm from the positive plate?_________ J

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Respuesta :

Answer:

a)  1.28 *10^5 N/C

b)2.05 *10^{-14} N

c) 4.83 *10^{-17} J

Explanation:

Given Data:

Distance between the plates, d = 4.67 mm

[tex]= (4.67) *10^{-3} m[/tex]

[/tex]= 4.67 *10^{-3} m[/tex]

Potential difference, V = 600 V

Solution:

(a) The  magnitude of the electric field between the plates is,

    [tex]E = \frac{V}{d}[/tex]  

[tex]= \frac{600 V}{4.67 *10^{-3}} m[/tex]

  [tex]= 1.28 *10^5 V/m or 1.28 *10^5N/C[/tex]  

(b) Force on electron btwn the plates is,

   F = q E

 [tex]= (1.6 *10^{-19} C) (1.28 *10^5N/C[/tex]

 [tex]= 2.05 *10^{-14} N[/tex]  

(c) Work done on the electron is

   W = F * s

 [tex]= (2.05 *10^{-14} N) * (5.31 *10^{-3} m - 2.95 *10^{-3} m)[/tex]

 [tex]= 4.83 *10^{-17} J[/tex]

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