The time before they met - 2

Determine the time for the charge q (mass m) touch the conducting plane as shown above if the time is t = ( π d ) x y ε 0 m q t=\frac{(\pi d)^{x}\sqrt{y\varepsilon_{0}m}}{q} find 2 x + y 2x +y


The answer is 5.

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1 solution

Theodore Yoong
Mar 1, 2014

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 1 4 π ε 0 q 2 ( 2 d ) 2 \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 μ = 1 1 m + 1 m = m 2 \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 m 2 \frac { m }{ 2 } crashing into some place 2d away under the influence of some pre-existing force.

By energy conservation, we obtain μ 2 ( d r d t ) 2 1 4 π ε 0 q 2 r = 1 4 π ε 0 q 2 2 d \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.

A very good solution but besides using conservation of energy we can use the concept of flattened ellipse to find the time along with the reduced mass concept. I used it and got the answer!

Pinak Wadikar - 7 years, 1 month ago

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Can you post that solution

Gauri shankar Mishra - 5 years, 1 month ago

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