The time before they met - 2

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 $\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }^{ 2 } }{ { (2d) }^{ 2 } }$ , making this essentially a central force problem. Thus, we denote the reduced mass as $\mu =\frac { 1 }{ \frac { 1 }{ m } +\frac { 1 }{ m } } =\frac { m }{ 2 }$ . That being said, we can now visualise the problem as some random body of mass $\frac { m }{ 2 }$ crashing into some place 2d away under the influence of some pre-existing force.

By energy conservation, we obtain $\frac { \mu }{ 2 } { \left( \frac { dr }{ dt } \right) }^{ 2 }-\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }^{ 2 } }{ r } =-\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }^{ 2 } }{ 2d }$ . Rearranging and solving the differential equation via integration by substitution, we obtain x=3/2 and y=2, giving the answer as 5.