Oppositely charged parallel plates are separated by 5.26 mm. A potential differe

Question

Oppositely charged parallel plates are separated by 5.26 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 2.90 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 2.90 mm from the positive plate?
3 J

Explanation / Answer

separation S = 5.26 * 10 ^ -3 m potential difference V = 600 V (a). the magnitude of the electric field strength between theplates E = V / S                                                                                                     = 114068.44 N /C (b).  the magnitude of the force on an electronbetween the plates F =E q where q = charge = 1.6 * 10 ^ -19 C So, F = 1.825 * 10 ^ -14 N (c). work done W = ( Eq ) S                             = 9.599 * 10 ^ -17 J

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