Question

A proton moves through a uniform electric field of 5.01 x 10 3 N/C. Calculate (a) the accelerationwith which the proton is moving and (b) the time taken by the proton to cover a distance of 4.8 cm.

Answer 1

a = eE/m = 1.6×10^-19× 5×10^3 / 1.6×10^-27
a = 5×10^11m/s^2

Answer 2

Explanation:

It is given that,

Electric field, $E=5.01\times 10^3\ N/C$

Charge of proton, $q=1.6\times 10^{-19}\ C$

Mass of proton, $m=1.67\times 10^{-27}\ kg$

(a) Electric force is given by :

$F=qE$

Also, F = ma

$a=\dfrac{qE}{m}$

$a=\dfrac{1.6\times 10^{-19}\ C\times 5.01\times 10^3\ N/C}{1.67\times 10^{-27}\ kg}$

$a=4.8\times 10^{11}\ m/s^2$

(b) Initially the proton is at rest, u = 0

Let v is the final velocity.

$s=ut+\dfrac{1}{2}at^2$

$t=\sqrt{\dfrac{2s}{a}}$

s is the displacement, s = 4.8 cm = 0.048 m

$t=\sqrt{\dfrac{2\times 0.048}{4.8\times 10^{11}}$

$t=4.4\times 10^{-7}\ s$

Hence, this is the required solution.