Oppositely charged parallel plates are separated by 5.31 mm. A potential differe

Question

Oppositely charged parallel plates are separated by 5.31 mm. A potential difference of 600 V existsbetween the plates. (a) What is the magnitude of the electric fieldstrength between the plates?
1 N/C
(b) What is the magnitude of the force on an electron between theplates?
2 N
(c) How much work must be done on the electron to move it to thenegative plate if it is initially positioned 2.89 mm from the positive plate?
3 J (a) What is the magnitude of the electric fieldstrength between the plates?
1 N/C
(b) What is the magnitude of the force on an electron between theplates?
2 N
(c) How much work must be done on the electron to move it to thenegative plate if it is initially positioned 2.89 mm from the positive plate?
3 J

Explanation / Answer

   we are given that    d = 5.31 mm       = 5.31 x 10-3m    V = 600 V (a)    V = E d    so the magnitude of the electric fieldstrength between the plates is given by    E = V / d        = ......... V /m (b)    the force is givern by    F = E q       = E (1.602 x10-19 C)       = ........... N (c)    the work that must be done is given by    W = q E d    here d will be    d = 5.31 mm - 2.89 mm       = .......... mm    W= ........... J

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