Separated by a distance of 2.00 cm , two large conducting plates are positioned parallel to one another. An electron released from rest near one plate travels to the opposite plate in 2.00 microseconds. Determine the surface charge density on the inner surfaces.

Step-by-step solution

Given the distance between the plates, d = 2.00 cm = 0.02 m , and the time taken by the electron, t = 2.00 s = 2.00 10⁻⁶ s . Using the kinematic equation for a particle starting from rest: d = 1 2 a t^2 Solving for acceleration a : a = 2d t^2 = 2 0.02 (2.00 10⁻⁶)^2 = 0.04 4.00 10⁻¹² = 10¹⁰ m/s ^2 The force experienced by the electron in the uniform electric field E is F = eE = m_e a . Thus, the electric field between the plates is: E = m_e a e = 9.11 10⁻³¹ 10¹⁰ 1.60 10⁻¹⁹ = 5.69 10⁻² V/m For two large parallel cond