Two facing surfaces of two large parallel conducting plates separated by 13.5 cm
ID: 2027034 • Letter: T
Question
Two facing surfaces of two large parallel conducting plates separated by 13.5 cm have uniform surface charge densities such that are equal in magnitude but opposite in sign. The difference in potential between the plates is 450 V.(b) What is the magnitude of the electric field between the plates?
kV/m
(c) An electron is released from rest next to the negatively charged surface. Find the work done by the electric field on the electron as the electron moves from the release point to the positive plate. Express your answer in both electron volts and joules.
eV
J
(d) What is the change in potential energy of the electron when it moves from the release point to the positive plate?
eV
(e) What is its kinetic energy when it reaches the positive plate?
eV
Explanation / Answer
GIVEN The potential difference between the plates is, v = 450 V The separation between the two plates, x = 13.5 cm = 0.135 m ----------------------------------------------------------------------------- b) The magnitude of electric field between the plates, E = v/x = 450 V /0.135 m = 3333.33 v/m = 3.333 kv/m ----------------------------------------------------------------------------- c) The work done by the electric field on the electron as the electron moves from the release point to the positive is given by the equation, W = q v W = 1.6 * 10^-19 * 450 V = 720 * 10^-19 J = 7.2 * 10^-17 J Therfore W = (7.2 * 10^-17 J) / (1.6 * 10^-19 ) = 450 eV ----------------------------------------------------------------------------- d) The change in potential energy, U = - W = - 450 eV ------------------------------------------------------------------------- (e) its kinetic energy K=-U= 450 eV GIVEN The potential difference between the plates is, v = 450 V The separation between the two plates, x = 13.5 cm = 0.135 m ----------------------------------------------------------------------------- b) The magnitude of electric field between the plates, E = v/x = 450 V /0.135 m = 3333.33 v/m = 3.333 kv/m ----------------------------------------------------------------------------- c) The work done by the electric field on the electron as the electron moves from the release point to the positive is given by the equation, W = q v W = 1.6 * 10^-19 * 450 V = 720 * 10^-19 J = 7.2 * 10^-17 J Therfore W = (7.2 * 10^-17 J) / (1.6 * 10^-19 ) = 450 eV ----------------------------------------------------------------------------- d) The change in potential energy, U = - W = - 450 eV ------------------------------------------------------------------------- (e) its kinetic energy K=-U= 450 eV (e) its kinetic energy K=-U= 450 eVRelated Questions
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