Answer to Question #287272 in Field Theory for Chaeyoung

Question

 A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.60 cm distant from the first, in a time interval of 3.20 x 10-6 s. 

(a) Find the magnitude of the electric field.

 (b) Find the speed of the proton when it strikes the negatively charged plate.

Answer 1

Solution

Time interval

t=3.20 x 10$^{-6}$ s

Distance between plates

d=1.6cm

Charge of proton

q=1.6$\times10^{-19}C$

Mass of proton

M=1.67$\times10^{-27}Kg$

a) so magnitude of the electric field

$E=\frac{2dM}{qt^2}\\E=\frac{2\times0.016\times1.67\times10^{-27}}{1.6\times10^{-19}(3.20\times10^{-6})^2}$

E=32.62V/m

b) now speed

$v=\frac{qEt}{m}\\=\frac{1.6\times10^{-19}\times32.62\times3.2\times10^{-6}}{1.67\times10^{-27}}\\=10000m/sec.$