Two parallel plates are in the shape of a square with dimension 26.9 cm by 26.9

Question

Two parallel plates are in the shape of  a square with dimension 26.9 cm by 26.9 cm, separated by 1.0 cm.

The left plate holds the total charge of -7.4 nC, the right one holds equal but positive charge.

A proton enters a region between the positively and negatively charged parallel plates through a hole in the left (negatively charged) one. It has the initial velocity of 271.9 km/s.

The proton traverses the distance between the plates, and exits through a hole in the positively charged one. What is its speed when it exits the region between the plates?

Explanation / Answer

area of plates of capacitor A = 26.9 cm x 26.9 cm = 723.61 cm^2 = 723.61 x 10^-4 m^2
for parallel plate capacitor , Q = 0 x E x A
7.4 nC = 0 x E x 723.61 x 10^-4
where , 0 = 8.854 × 10^12 F/m
E = (7.4 x 10^-9)/ (8.854 × 10^12 x 723.61 x 10^-4 ) = 1.155 x 10^-4 N/C
    
now,
F=qE
and
F=ma (Newton's second law)


Equating above two equations gives: ma=qE
a =qE/m

q and m are constants so the ''a' vector is the same as the E vector to be balanced both sides of the equ with the correct sign.
charge of proton = 1.6 x 10^-19 C
a =qE/m = ( 1.6 x 10^-19 x 1.155 x 10^-4)/ (1.67 x 10^-27) = 1.1066 x 10^4 m/s^2
now initial velocity , u = 271.9 km/s = 271.9 x 10^3 m/s
let final velocity be v.
distance travelled , d = 1 cm = 0.01 m
using relation , v^2 = u^2 + 2 a x d = (271.9 x 10^3)^2 + 2*1.1066 x 10^4 x 0.01 = 7.393 x 10^10
v = 271900 m/s = 271.9 km/s

here v = u as E is quite weak.

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