Question

A proton initially at rest is accelerated by a uniform electric field. The proton moves 4.76 cm in 1.10 x 10^-6 s. Find the voltage drop through which the proton moves.?

Answer 1

Answer:

The voltage drop through which the proton moves is 39.1 V.

Explanation:

Given that,

Distance = 4.76 cm

Time $t=1.10\times10^{-6}\ s$

We need to calculate the acceleration

Using equation of motion

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

Where, s = distance

a = acceleration

t = time

Put the value in the equation

$4.76\times10^{-2}=\dfrac{1}{2}\times a \times(1.10\times10^{-6})^2$

$a=\dfrac{2\times 4.76\times10^{-2}}{(1.10\times10^{-6})^2}$

$a=7.87\times10^{10}\ m/s^2$

We need to calculate the voltage drop

Using formula of electric field

$F=qE$

$F = q\dfrac{V}{d}$....(I)

Using newton's second law

$F = ma$....(II)

Put the value of F in equation (I) from equation (II)

$ma=\dfrac{qV}{d}$

$V=\dfrac{mad}{q}$

Where, q = charge

a = acceleration

d = distance

m= mass of proton

Put the value into the formula

$V=\dfrac{1.67\times10^{-27}\times7.87\times10^{10}\times4.76\times10^{-2}}{1.6\times10^{-19}}$

$V=39.1\ V$

Hence, The voltage drop through which the proton moves is 39.1 V.