Oppositely charged parallel plates are separated by5.35 mm. A potential differen

Question

Oppositely charged parallel plates are separated by5.35 mm. A potential difference of 600V exists between the plates (a) What is the magnitude of the electric field strengthbetween the plates? (b) What is the magnitude of the force on an electron betweenthe plates?
2 N
(c) How much work must be done on the electron to move it to thenegative plate if it is initially positioned 2.90 mm from the positive plate?
3 J
Oppositely charged parallel plates are separated by5.35 mm. A potential difference of 600V exists between the plates (a) What is the magnitude of the electric field strengthbetween the plates? (b) What is the magnitude of the force on an electron betweenthe plates?
2 N
(c) How much work must be done on the electron to move it to thenegative plate if it is initially positioned 2.90 mm from the positive plate?
3 J

Explanation / Answer

      Given           distancebetween two plates   d = 5.35 mm   =0.00535 m           potentialbetween two plates V = 600 V              a) magnitude of electric fieldstrength   E = V/ d                                                                    = 600 / 0.00535                                                                  = 1.12*105 N / m b) magnitude of force on electron    F = E q              q = charge of electron   = 1.6*10-19C                      there fore F =   1.12*105 *1.6*10-19                                          = 1.792*10-14 N       c)

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