A uniform electric field exists in the region between twooppositely charged para

Question

A uniform electric field exists in the region between twooppositely charged parallel plates 1.58 cm apart.
A proton is released from rest at the surface of the positivelycharged plate and strikes the surface of the opposite plate in atime interval 1.57×106 s .(Use1.60×1019 C for the magnitude of thecharge on an electron and 1.67×1027 kgfor the mass of a proton.) Find the magnitude of the electricfield  (-----do i need to integrate here?-------) Find the speed of the proton at the moment itstrikes the negatively charged plate. A uniform electric field exists in the region between twooppositely charged parallel plates 1.58 cm apart.
A proton is released from rest at the surface of the positivelycharged plate and strikes the surface of the opposite plate in atime interval 1.57×106 s .(Use1.60×1019 C for the magnitude of thecharge on an electron and 1.67×1027 kgfor the mass of a proton.) Find the magnitude of the electricfield  (-----do i need to integrate here?-------) Find the speed of the proton at the moment itstrikes the negatively charged plate. .(Use1.60×1019 C for the magnitude of thecharge on an electron and 1.67×1027 kgfor the mass of a proton.) Find the magnitude of the electricfield  (-----do i need to integrate here?-------) Find the speed of the proton at the moment itstrikes the negatively charged plate.

Explanation / Answer

d = (1/2) at2 a = 2d /t2 = 2 * 0.0158 / ( 1.57 x10-6 )2 =                  =   1.2820 x 1010 m/s2 Then we know that   Force=mass * acc
So we can find the force on the proton
F= 1.67 x 10-27 * 1.2820 x1010 = 2.14 x 10-17 Newtons
Then we know that    F = q E  so E = m a / q  =  1.67 x 10-27 x 1.2820 x1010 / 1.60 x 10-19  =                  = 133.8 N/C isthe strength of the electric field.
Speed can be found by using
v = acc * time =   1.2820 x1010 * 1.57 x 10-6 = 20127.4 m/s
HAHAHAHAHAHAHA

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