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प्रश्न
Two large conducting plates are placed parallel to each other with a separation of 2⋅00 cm between 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.
उत्तर
Distance travelled by the electron, d= 2 cm
Time taken to cross the region, t = 2× 10-6 s
Let the surface charge density at the conducting plates be σ.
Let the acceleration of the electron be a.
Applying the 2nd equation of motion, we get:
`"d" = 1/2 "a""t"^2`
`=> "a" = (2"d")/"t"^2`
This acceleration is provided by the Coulombic force. So,
`"a" = "qE"/m = "2d"/"t"^2`
`=> "E" = (2"md")/"qt"^2`
`"E" = (2 xx (9.1xx10^-31) xx (2 xx 10^-2))/((1.6 xx 10^-19) xx ( 4 xx 10^-12)`
E = 5.6875 × 10-2 N/C
Also, we know that electric field due to a plate,
`"E" = sigma/∈_0`
⇒ σ =∈_0 E
⇒ σ = ( 8.85 × 10-12) × ( 5.68 × 10-2 ) C/m2
⇒ σ = 50.33 × 10-14 C/m2 = 0.503 × 10-12 C/m2
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