Two large conducting plates are placed parallel to each other with a separation of 2.00 cm betweeen them. An electron starting from rest near one of the plates reaches the other plate in 2.00 microseconds. Find the surface charge density on the inner surfaces.

Explanation

` d = 2.00 cm = 2 xx (10^-2)m, `
` t = 2.0 xx (10^-6)s `
` Here d= `1/2 at^2` :.a=2d/t^2 `
` or a = qE/m = 2d/t^2 `
` or E= 2md/qt^2`
` = (2 xx 9.1 xx (10^-31) xx 2 xx (10^2) / 1.6 xx (10^19) xx 4 xx (10^-12)) `
` = 5.6875 xx (10^-2) N/C`
` E = sigma / epsilon_0`
` sigma = 8.9 xx (10^-12) xx 5.6875 xx (10^-2) `
` = 0.505 xx (10^-12)C/(m^2).`