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Determine the time for the charge q (mass m) touch the conducting plane as shown above if the time is
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Denote the conducting plane to be the x-y plane, and orthogonal to it the z-axis, in which the actual charge lies on, i.e. its position co-ordinates are (0, 0, d). By the first uniqueness theorem, we can replace the grounded plate with an image point charge of -q a distance 2d away from the actual charge of +q as it allows for the potential along the x-y plane to remain zero. Simply put, the image charge has charge -q and co-ordinates (0, 0, -d).
The only (non-negligible) force of interaction is that of the Coloumbic attraction between the 2 charges or magnitude 4 π ε 0 1 ( 2 d ) 2 q 2 , making this essentially a central force problem. Thus, we denote the reduced mass as μ = m 1 + m 1 1 = 2 m . That being said, we can now visualise the problem as some random body of mass 2 m crashing into some place 2d away under the influence of some pre-existing force.
By energy conservation, we obtain 2 μ ( d t d r ) 2 − 4 π ε 0 1 r q 2 = − 4 π ε 0 1 2 d q 2 . Rearranging and solving the differential equation via integration by substitution, we obtain x=3/2 and y=2, giving the answer as 5.