A uniform electric field of magnitude 628 N/C exists between two parallel plates

Question

A uniform electric field of magnitude 628 N/C exists between two parallel plates that are 4.20 cm apart. A proton is released from rest at the positive plate at the same instant an electron is released from rest at the negative plat

A. Determine the distance from the positive plate at which the two pass each other. Ignore the electrical attraction between the proton and electron.

B. Repeat part (a) for a sodium ion (Na+) and a chloride ion (Cl?).

Need Explaination!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! or no points....

Explanation / Answer

A) By F = qE
=>F = eE = [1.6 x 10^-19] x 628 = 1.0048 x 10^-16 N
Thus by a = F/m
=>a(electron) = [1.0048 x 10^-16/9.11 x 10^-31] = 1.103 x 10^14 m/s^2

& a(proton) = [1.0048 x 10^-16/1.67 x 10^-27] = 6.017 x 10^10 m/s^2

Let the two met after t sec
By
s = ut + 1/2at^2

=>s(p) = 0 + 1/2* 6.017* 10^10 * t^2 -------------(i)

& s(e) = 0 + 1/2 * 1.103*10^14 * t^2 -------------(ii)

=>By (i) + (ii) :-

=>4.2 x 10^-2 = 1/2 * 10^10 * t^2 [6.017 + 11030]

=>
=>t = 2.759 x 10^-8 sec

(a) distance form positive plat e,
s(p) = 1/2 * 6.017 * 10^10 * (2.759 x 10^-8)^2 = 2.29 x 10^-6 m

B)

No w,
mass of sodium ion = 22.99 g / 6.022*10^23 atoms = 3.818 * 10^ -26 kg
charge = 1.6*10^-19 +ve

similarly ,
mass of Chlorine ion = 35.5 g / 6.022*10^23 atoms = 5.895 * 10^ -26 kg
charge = 1.6*10^-19 -ve

so,,
similar to above calculation ,
By F = qE
=>F = eE = [1.6 x 10^-19] x 628 = 1.0048 x 10^-16 N
Thus by a = F/m
=>a(Na+) = [1.0048 x 10^-16/ 3.818 * 10^ -26] = 2.632 x 10^9m/s^2

& a(Cl-) = [1.0048 x 10^-16/ 5.895 * 10^ -26] =1.7045 x 10^9 m/s^2

Let the two met after t sec
By
s = ut + 1/2at^2

=>s(Na+) = 0 + 1/2* 2.632 x 10^9 * t^2 -------------(i)

& s(Cl-) = 0 + 1/2 * 1.7045 x 10^9 * t^2 -------------(ii)

=>By (i) + (ii) :-

=>4.2 x 10^-2 = 1/2 * 10^9 * t^2 [2.632+1.7045]

=>t =4.4012*10^-6 sec

so,
b)
distance form positive plate,
s(Na+) = 1/2 * 2.632 x 10^9 * (4.4012*10^-6)^2 =0.02549 m
so ,
d = 2.55 cm (approx)

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